Intermediate Algebra Practice Questions — ACT — Page 11

Question 201Question →

For the functions $f(x) = (x - 2)^2 - 5$ and $g(x) = |x + 1| - 3$ , what is the value of $f(g(-2))$ ?

9

-5

11

31

-22

Show answer & explanation

Answer: 11

Answer

11

To find the value of

f(g(-2))

, first evaluate the inner function

g(x)

at

x = -2

. Substituting

-2

into

g(x) = |x + 1| - 3

gives

g(-2) = |-2 + 1| - 3 = 1 - 3 = -2

. Next, substitute this output as the input for the outer function

f(x)

. Substituting

-2

into

f(x) = (x - 2)^2 - 5

yields

f(-2) = (-2 - 2)^2 - 5 = (-4)^2 - 5 = 16 - 5 = 11

.

Step-by-Step Solution

1

Evaluate the inner function at the given input value

g(-2) = |-2 + 1| - 3 = |-1| - 3 = 1 - 3 = -2

In function composition

f(g(x))

, the inner function must be evaluated first to determine the input for the outer function.

2

Evaluate the outer function using the output of the inner function

f(-2) = (-2 - 2)^2 - 5 = (-4)^2 - 5 = 16 - 5 = 11

Substitute the result from the inner function evaluation into the outer function to find the final value.

Key Concept

Function composition and evaluation

Estimated Time:1m 0s

Question 202Question →

If $x$ is a real number greater than 2 such that $\log_2(x^2 - 4) - \log_2(x - 2) = 3$ , what is the value of $x$ ?

1

4

6

7

16

Show answer & explanation

Answer: 6

Answer

The value of

x

is 6.

Applying the quotient property of logarithms simplifies the equation to

\log_2\left(\frac{x^2 - 4}{x - 2}\right) = 3

. Factoring the numerator as

(x-2)(x+2)

and canceling the common factor

(x-2)

yields

\log_2(x+2) = 3

. Rewriting this in exponential form gives

x + 2 = 2^3 = 8

, which yields

x = 6

.

Step-by-Step Solution

1

Combine the logarithmic terms on the left side of the equation using the quotient property of logarithms:

\log_b A - \log_b B = \log_b\left(\frac{A}{B}\right)

.

\log_2\left(\frac{x^2 - 4}{x - 2}\right) = 3

Consolidating the two logarithmic terms simplifies the equation into a single logarithmic term.

2

Factor the numerator as a difference of squares and simplify the rational expression.

\log_2(x + 2) = 3

(since

x > 2

,

x - 2 \neq 0

)

Factoring

x^2 - 4

as

(x - 2)(x + 2)

allows the cancellation of the common factor

(x - 2)

.

3

Convert the logarithmic equation into its equivalent exponential form:

\log_b Y = c \Rightarrow Y = b^c

.

x + 2 = 2^3

This removes the logarithm and sets up a linear equation to solve.

4

Evaluate the exponent and solve for

x

.

x + 2 = 8 \Rightarrow x = 6

Evaluating

2^3

as 8 and subtracting 2 from both sides isolates

x

.

Key Concept

Solving logarithmic equations using properties of logarithms and algebraic factoring

Estimated Time:1m 30s

Question 203Question →

Which of the following is the set of all real values of $p$ for which the inequality $|7 - 3p| \ge 14$ is true?

-\frac{7}{3} \le p \le 7

p \le -\frac{7}{3} \text{ or } p \ge 7

p \le -\frac{7}{3}

p \le -\frac{7}{2} \text{ or } p \ge \frac{7}{2}

p \ge -\frac{7}{3} \text{ or } p \le 7

Show answer & explanation

Answer: $p \le -\frac{7}{3} \text{ or } p \ge 7$

Answer

p \le -\frac{7}{3} \text{ or } p \ge 7

To solve the inequality

|7 - 3p| \ge 14

, we split it into two separate inequalities:

7 - 3p \ge 14

or

7 - 3p \le -14

. Solving the first inequality gives

-3p \ge 7

, which simplifies to

p \le -\frac{7}{3}

after dividing by

-3

and reversing the inequality sign. Solving the second inequality gives

-3p \le -21

, which simplifies to

p \ge 7

after dividing by

-3

and reversing the inequality sign. Combining these two cases yields the set of values

p \le -\frac{7}{3}

or

p \ge 7

.

Step-by-Step Solution

1

Set up the two compound inequalities representing the absolute value inequality

|7 - 3p| \ge 14

.

7 - 3p \ge 14

or

7 - 3p \le -14

An absolute value inequality of the form

|u| \ge c

(where

c > 0

) is equivalent to the union of

u \ge c

or

u \le -c

.

2

Solve the first inequality

7 - 3p \ge 14

.

p \le -\frac{7}{3}

Subtract

7

from both sides to get

-3p \ge 7

. Then, divide by

-3

and reverse the inequality sign because of division by a negative number.

3

Solve the second inequality

7 - 3p \le -14

.

p \ge 7

Subtract

7

from both sides to get

-3p \le -21

. Then, divide by

-3

and reverse the inequality sign because of division by a negative number.

4

Combine the two case solutions to state the final solution set.

p \le -\frac{7}{3} \text{ or } p \ge 7

The complete solution set is the union of the solutions from both individual cases.

Key Concept

Solving absolute value inequalities of the form

|ax + b| \ge c

Question 204Question →

If $a > 1.5$ is a real number that satisfies the equation $4^a - 3(2^{a+1}) + 8 = 0$ , what is the value of $a$ ?

Show answer & explanation

Answer: 2

Answer

The correct answer is 2.

Substituting

u = 2^a

turns the equation into

u^2 - 6u + 8 = 0

. Factoring gives

(u-2)(u-4)=0

, meaning

u=2

or

u=4

. Reversing the substitution gives

2^a = 2 \implies a=1

and

2^a = 4 \implies a=2

. Since

a > 1.5

, the correct value is 2.

Step-by-Step Solution

1

Express the equation in terms of base 2.

(2^a)^2 - 6(2^a) + 8 = 0

Since 4 is

2^2

and

2^{a+1} = 2 \cdot 2^a

, expressing all terms in base 2 allows for algebraic substitution.

2

Substitute

u = 2^a

to form a quadratic equation.

u^2 - 6u + 8 = 0

Substitution simplifies the exponential equation into a standard quadratic form.

3

Factor the quadratic equation.

(u-2)(u-4) = 0

Factoring allows us to find the roots of the quadratic equation.

4

Solve for the variable

a

.

a = 1

or

a = 2

Solving

2^a = 2

yields

a = 1

, and solving

2^a = 4

yields

a = 2

.

5

Apply the given constraint on

a

.

a = 2

The problem states that

a > 1.5

, so

a = 1

is discarded and

a = 2

is the correct value.

Key Concept

Solving exponential equations using quadratic substitution

Alternative Method

Instead of using substitution, test values for

a

. Since

a > 1.5

, testing small integer values starting with

a=2

shows

4^2 - 3(2^3) + 8 = 16 - 24 + 8 = 0

, validating that

a=2

is the solution.

Estimated Time:1m 30s

Question 205Question →

What is the complete solution set, expressed in interval notation, for the inequality $\left| \frac{2-3t}{4} \right| > 2$ ?

(-\infty, -2)

\left(-2, \frac{10}{3}\right)

(-\infty, -8) \cup (8, \infty)

(-\infty, -2) \cup \left(\frac{10}{3}, \infty\right)

\left(-\infty, -\frac{10}{3}\right) \cup (2, \infty)

Show answer & explanation

Answer: $(-\infty, -2) \cup \left(\frac{10}{3}, \infty\right)$

Answer

The complete solution set is

(-\infty, -2) \cup \left(\frac{10}{3}, \infty\right)

The correct answer shows

(-\infty, -2) \cup \left(\frac{10}{3}, \infty\right)

because solving the absolute value inequality

|2-3t| > 8

requires splitting it into two inequalities:

2-3t > 8

and

2-3t < -8

. Solving the first yields

t < -2

after reversing the inequality sign when dividing by

-3

. Solving the second yields

t > \frac{10}{3}

after also reversing the inequality sign. The union of these two intervals is

(-\infty, -2) \cup \left(\frac{10}{3}, \infty\right)

.

Step-by-Step Solution

1

Multiply both sides of the inequality by

4

to isolate the absolute value term.

|2-3t| > 8

Eliminating the denominator simplifies the absolute value expression.

2

Split the absolute value inequality into two separate linear inequalities representing the positive and negative cases.

2-3t > 8

or

2-3t < -8

An absolute value greater than a positive number

c

is equivalent to the expression being greater than

c

or less than

-c

.

3

Solve the first inequality: subtract

2

from both sides, then divide by

-3

and reverse the inequality sign.

-3t > 6 \implies t < -2

Dividing an inequality by a negative number requires flipping the inequality sign.

4

Solve the second inequality: subtract

2

from both sides, then divide by

-3

and reverse the inequality sign.

-3t < -10 \implies t > \frac{10}{3}

Dividing an inequality by a negative number requires flipping the inequality sign.

5

Combine the two solutions using interval notation.

(-\infty, -2) \cup \left(\frac{10}{3}, \infty\right)

The union of the two intervals represents the complete set of values that satisfy either inequality.

Key Concept

Solving absolute value inequalities of the form

|ax+b| > c

by splitting them into two cases and reversing the inequality sign when dividing by a negative number.

Estimated Time:1m 30s

Question 206Question →

For all real numbers $x$ and $y$ such that $x > y > 0$ , which of the following expressions is equivalent to $\log_5(x^2 - y^2) - \log_5(x - y)$ ?

\log_5(x-y)

\log_5(x^2 - y^2 - x + y)

\log_5(x) - \log_5(y)

\log_5(x+y)

x + y

Show answer & explanation

Answer: $\log_5(x+y)$

Answer

The expression is equivalent to \log_5(x + y)

The correct answer is found by using the quotient property of logarithms to combine the terms:

\log_5(x^2 - y^2) - \log_5(x - y) = \log_5\left(\frac{x^2 - y^2}{x - y}\right)

. Factoring the difference of squares in the numerator yields

\log_5\left(\frac{(x-y)(x+y)}{x-y}\right)

. Canceling the common factor

(x-y)

simplifies the expression to

\log_5(x+y)

.

Step-by-Step Solution

1

Combine the logarithms using the quotient property of logarithms:

\log_b(A) - \log_b(B) = \log_b\left(\frac{A}{B}\right)

.

\log_5\left(\frac{x^2 - y^2}{x - y}\right)

This simplifies the difference between two logarithms with the same base into a single logarithm.

2

Factor the difference of squares in the numerator:

x^2 - y^2 = (x - y)(x + y)

.

\log_5\left(\frac{(x - y)(x + y)}{x - y}\right)

This allows for the cancellation of common factors in the fraction.

3

Cancel the common factor

(x - y)

from both the numerator and the denominator.

\log_5(x + y)

Since

x > y > 0

, the term

x - y

is non-zero, making the division valid.

Key Concept

Quotient property of logarithms and difference of squares factoring

Alternative Method

Instead of applying the quotient property first, one can factor the argument of the first term using the difference of squares:

\log_5(x^2 - y^2) = \log_5((x - y)(x + y))

. Next, apply the product property of logarithms to split this term:

\log_5(x - y) + \log_5(x + y)

. Substituting this back into the original expression gives

\log_5(x - y) + \log_5(x + y) - \log_5(x - y)

, which simplifies directly to

\log_5(x + y)

.

Estimated Time:1m 0s

Question 207Question →

If $\log_2(x + 5) + \log_2(x - 1) = 4$ , what is the value of $x$ ?

Show answer & explanation

Answer: 3

Answer

The correct answer is 3.

The correct answer is 3. Combining the logarithmic terms using the product property gives

\log_2((x + 5)(x - 1)) = 4

. Converting this to exponential form yields

(x + 5)(x - 1) = 2^4 = 16

. Expanding and rewriting in standard form gives

x^2 + 4x - 21 = 0

, which factors as

(x + 7)(x - 3) = 0

. This gives potential solutions of

x = 3

and

x = -7

. However,

x = -7

results in negative arguments for the logarithms in the original equation, making it extraneous. Thus, the only valid solution is 3.

Step-by-Step Solution

1

Apply the product property of logarithms to combine the terms on the left side of the equation.

\log_2((x + 5)(x - 1)) = 4

The sum of logarithms with the same base is equal to the logarithm of their product:

\log_b(A) + \log_b(B) = \log_b(AB)

.

2

Convert the logarithmic equation to its equivalent exponential form.

(x + 5)(x - 1) = 2^4

By definition,

\log_b(Y) = C

is equivalent to

b^C = Y

.

3

Expand the product and simplify the equation into standard quadratic form.

x^2 + 4x - 21 = 0

Expanding

(x + 5)(x - 1)

yields

x^2 + 4x - 5

. Subtracting 16 from both sides gives the standard quadratic form

Ax^2 + Bx + C = 0

.

4

Factor the quadratic equation to find the potential values of

x

.

(x + 7)(x - 3) = 0

, so

x = -7

or

x = 3

We need two numbers that multiply to

-21

and add to

4

, which are

7

and

-3

.

5

Check the potential solutions against the domain of the original logarithmic equation.

x = 3

The arguments of the logarithms,

x + 5

and

x - 1

, must be strictly positive. For

x = -7

, the arguments are negative, so

x = -7

is an extraneous solution. For

x = 3

, both arguments are positive, so

x = 3

is the only valid solution.

Key Concept

Solving logarithmic equations by applying logarithmic properties and converting to exponential form, while checking for extraneous solutions.

Question 208Question →

A circle in the standard $(x, y)$ coordinate plane is defined by the equation $(x - 1)^2 + (y - 2)^2 = 10$ , and a line is defined by the equation $y = 3x - 1$ . The line intersects the circle at two points, $(x_1, y_1)$ and $(x_2, y_2)$ . What is the sum of the $y$ -coordinates of these two points of intersection?

Show answer & explanation

Answer: 4

Answer

The sum of the

y

-coordinates of the intersection points is 4.

Substituting

y = 3x - 1

into the circle's equation gives

(x-1)^2 + (3x-3)^2 = 10

. Factoring out 3 from the second term yields

(x-1)^2 + 9(x-1)^2 = 10

, which simplifies to

10(x-1)^2 = 10

, or

(x-1)^2 = 1

. Solving for

x

gives

x = 0

and

x = 2

. Substituting these values into the linear equation gives the corresponding

y

-coordinates:

y = -1

when

x = 0

, and

y = 5

when

x = 2

. The sum of these

y

-coordinates is

5 + (-1) = 4

.

Step-by-Step Solution

1

Substitute the expression for

y

from the linear equation into the circle's equation.

(x-1)^2 + (3x-3)^2 = 10

To reduce the system of two equations to a single equation in terms of

x

.

2

Factor out 3 from the term

(3x-3)

and simplify the equation.

10(x-1)^2 = 10

, which simplifies to

(x-1)^2 = 1

To solve for the

x

-coordinates of the intersection points.

3

Solve the simplified quadratic equation for

x

.

x_1 = 2

and

x_2 = 0

To find the

x

-coordinates of the two intersection points.

4

Substitute the

x

-values back into the linear equation

y = 3x - 1

to find the corresponding

y

-coordinates.

y_1 = 3(2) - 1 = 5

and

y_2 = 3(0) - 1 = -1

To determine the

y

-coordinates of the intersection points

(2, 5)

and

(0, -1)

.

5

Add the two

y

-coordinates together.

5 + (-1) = 4

To find the sum of the

y

-coordinates as requested by the question.

Key Concept

Solving systems of linear and circular equations by substitution

Alternative Method

Instead of finding the individual coordinates, substitute

y_1 = 3x_1 - 1

and

y_2 = 3x_2 - 1

to write the sum as

y_1 + y_2 = 3(x_1 + x_2) - 2

. Expanding the substitution equation gives

10x^2 - 20x = 0

. By Vieta's formulas, the sum of the roots

x_1 + x_2 = -(-20)/10 = 2

. Substituting this back gives

y_1 + y_2 = 3(2) - 2 = 4

.

Estimated Time:1m 30s

Question 209Question →

A system of equations consists of the linear equation $y = x - 2$ and the quadratic equation $y = (x - 3)^2 - 5$ . If $(x_1, y_1)$ and $(x_2, y_2)$ are the two distinct real solutions to this system, what is the value of the product $y_1 y_2$ ?

-10

-4

3

4

6

Show answer & explanation

Answer: -4

Answer

The product of the two

y

-coordinates of the intersection points is

-4

.

Substituting the linear expression for

y

into the quadratic equation yields the quadratic equation

x^2 - 7x + 6 = 0

. Factoring this equation gives the

x

-coordinates

6

and

1

. Substituting these values back into the linear equation

y = x - 2

yields the corresponding

y

-coordinates

4

and

-1

. The product of these

y

-coordinates is

4 \times (-1) = -4

.

Step-by-Step Solution

1

Equate the two expressions for

y

to solve for the

x

-coordinates of the intersection points.

x - 2 = (x - 3)^2 - 5

At the points of intersection, the

y

-values of both equations are equal.

2

Expand the squared binomial and simplify the equation into standard quadratic form.

x^2 - 7x + 6 = 0

Expanding

(x - 3)^2

yields

x^2 - 6x + 9

. Substituting this back gives

x - 2 = x^2 - 6x + 4

. Subtracting

x

and adding

2

to both sides yields the standard quadratic equation.

3

Factor the quadratic equation to find the solutions for

x

.

x = 6

or

x = 1

The equation factors as

(x - 6)(x - 1) = 0

, giving the two

x

-values.

4

Substitute each

x

-value back into the linear equation

y = x - 2

to find the corresponding

y

-coordinates.

y_1 = 4

and

y_2 = -1

For

x = 6

,

y = 6 - 2 = 4

. For

x = 1

,

y = 1 - 2 = -1

.

5

Calculate the product of the two

y

-coordinates.

y_1 y_2 = 4 \times (-1) = -4

The question asks for the product of the two

y

-coordinates.

Key Concept

Solving a system of linear and quadratic equations using substitution and factoring.

Alternative Method

Alternatively, you can expand the quadratic equation first to

y = x^2 - 6x + 4

and set it equal to the linear equation

y = x - 2

. Solving for

x

yields

x^2 - 7x + 6 = 0

, from which you can find the coordinates and calculate their product.

Estimated Time:1m 30s

Question 210Question →

If $x$ is a real number such that $\log_4(x - 3) = \frac{1}{2} + \log_4(2)$ , what is the value of $x$ ?

Show answer & explanation

Answer: 7

Answer

The value of

x

is 7.

By moving the logarithmic terms to the same side and applying the quotient rule, the equation simplifies to

\log_4\left(\frac{x-3}{2}\right) = \frac{1}{2}

. Converting this to exponential form yields

\frac{x-3}{2} = 4^{1/2}

. Since

4^{1/2} = 2

, we have

\frac{x-3}{2} = 2

, which gives

x - 3 = 4

, or

x = 7

. Substituting

x = 7

back into the original equation shows it is a valid solution.

Step-by-Step Solution

1

Group logarithmic terms on one side of the equation.

\log_4(x - 3) - \log_4(2) = \frac{1}{2}

Grouping the logarithms allows them to be combined using logarithmic properties.

2

Apply the quotient property of logarithms.

\log_4\left(\frac{x - 3}{2}\right) = \frac{1}{2}

The quotient property states that

\log_b(a) - \log_b(c) = \log_b(\frac{a}{c})

.

3

Convert the equation from logarithmic form to exponential form.

\frac{x - 3}{2} = 4^{1/2}

A logarithmic equation

\log_b(y) = z

is equivalent to

b^z = y

.

4

Evaluate the fractional exponent and solve the linear equation for

x

.

x = 7

Since

4^{1/2} = 2

, the equation becomes

\frac{x - 3}{2} = 2

. Multiplying both sides by 2 gives

x - 3 = 4

, so adding 3 to both sides yields

x = 7

.

Key Concept

Solving logarithmic equations using properties of logarithms

Estimated Time:1m 30s

Question 211Question →

A system of equations consists of the linear equation $y = x + 5$ and the quadratic equation $y = (x - 3)^2 + 2$ . If the graphs of these two equations intersect at two distinct points, $(x_1, y_1)$ and $(x_2, y_2)$ , in the standard $(x, y)$ coordinate plane, what is the value of $y_1 + y_2$ ?

3

7

11

15

17

Show answer & explanation

Answer: 17

Answer

17

To find the points of intersection, we set the two equations equal:

x + 5 = (x - 3)^2 + 2

. Expanding the right side results in

x + 5 = x^2 - 6x + 9 + 2

, which simplifies to

x + 5 = x^2 - 6x + 11

. Subtracting

x + 5

from both sides gives the quadratic equation

x^2 - 7x + 6 = 0

. Factoring this equation yields

(x - 1)(x - 6) = 0

, giving

x

-coordinates of

1

and

6

. Substituting these values into the linear equation

y = x + 5

gives the

y

-coordinates:

y = 1 + 5 = 6

and

y = 6 + 5 = 11

. The sum of these

y

-coordinates is

6 + 11 = 17

.

Step-by-Step Solution

1

Expand the quadratic equation to its standard polynomial form.

y = x^2 - 6x + 11

Expanding

(x - 3)^2 + 2

allows for easier substitution and algebraic manipulation.

2

Equate the linear and quadratic expressions for

y

to find the intersection points.

x + 5 = x^2 - 6x + 11

Setting the two equations equal to each other identifies the

x

-coordinates where the graphs intersect.

3

Rearrange the equation into standard quadratic form and solve for

x

by factoring.

x^2 - 7x + 6 = 0 \implies (x - 1)(x - 6) = 0

, which gives

x = 1

and

x = 6

.

Factoring allows us to find the roots of the quadratic equation representing the intersection

x

-coordinates.

4

Substitute the

x

-coordinates back into the linear equation

y = x + 5

to determine the corresponding

y

-coordinates.

For

x = 1

,

y = 6

. For

x = 6

,

y = 11

.

Finding the

y

-coordinates corresponding to the intersection points.

5

Calculate the sum of the two

y

-coordinates.

6 + 11 = 17

The question asks specifically for the value of

y_1 + y_2

.

Key Concept

Solving systems of linear and quadratic equations by substitution and factoring

Question 212Question →

A temperature control system in a chemical reactor regulates target output temperature, in degrees Celsius, based on the reactor's pressure coefficient. The pressure coefficient $P$ is determined by the elapsed time $t$ , in minutes, according to the function $P(t) = (t - 3)^2 - 5$ . The target temperature $T$ is determined by the pressure coefficient $P$ according to the function $T(P) = 3|P - 2| + 1$ . What is the target output temperature of the system, in degrees Celsius, at time $t = 1$ minute?

-8

12

10

-4

46

Show answer & explanation

Answer: 10

Answer

10

The correct temperature is 10. By first finding the pressure at

t = 1

to be

P(1) = (1 - 3)^2 - 5 = 4 - 5 = -1

, and then substituting this pressure into the temperature function, we get

T(-1) = 3|-1 - 2| + 1 = 3(3) + 1 = 10

.

Step-by-Step Solution

1

Calculate the pressure coefficient

P(1)

at time

t = 1

minute by substituting

t = 1

into

P(t) = (t - 3)^2 - 5

.

P(1) = -1

Before calculating the temperature, we must find the input value

P

corresponding to

t = 1

.

2

Evaluate the target temperature

T(P)

at the pressure coefficient

P = -1

by substituting

P = -1

into

T(P) = 3|P - 2| + 1

.

T(-1) = 10

Applying the absolute value definition

|-3| = 3

, we find

3(3) + 1 = 10

.

Key Concept

Evaluating composite functions in a real-world scenario.

Alternative Method

Instead of evaluating step-by-step numerically, we can write the composite function symbolically:

T(P(t)) = 3|((t - 3)^2 - 5) - 2| + 1 = 3|(t - 3)^2 - 7| + 1

. Plugging in

t = 1

yields

T(P(1)) = 3|(1 - 3)^2 - 7| + 1 = 3|4 - 7| + 1 = 3|-3| + 1 = 10

.

Estimated Time:1m 30s

Question 213Question →

If $b$ is a positive real number unequal to $1$ such that $\log_b 3 = x$ and $\log_b 5 = y$ , what is the value of $\log_b \left( \frac{45}{b^2} \right)$ in terms of $x$ and $y$ ?

2x + y - 2

x^2 + y - 2

\frac{2x + y}{2}

2x + y - 2b

2x - y - 2

Show answer & explanation

Answer: $2x + y - 2$

Answer

2x + y - 2

The correct answer is

2x + y - 2

. By applying the quotient property of logarithms,

\log_b \left( \frac{45}{b^2} \right)

is rewritten as

\log_b 45 - \log_b(b^2)

. Factoring

45

as

3^2 \cdot 5

allows the first term to be expanded using the product and power properties into

2\log_b 3 + \log_b 5

. Simplifying

\log_b(b^2)

to

2

and substituting

x

and

y

yields

2x + y - 2

.

Step-by-Step Solution

1

Apply the quotient property of logarithms.

\log_b \left( \frac{45}{b^2} \right) = \log_b 45 - \log_b(b^2)

The logarithm of a quotient is the difference of the logarithms of the numerator and the denominator:

\log_b(M/N) = \log_b M - \log_b N

.

2

Factor the number 45 and apply the product property of logarithms.

\log_b(3^2 \cdot 5) - \log_b(b^2) = \log_b(3^2) + \log_b 5 - \log_b(b^2)

Since

45 = 9 \cdot 5 = 3^2 \cdot 5

, we can use the product property:

\log_b(M \cdot N) = \log_b M + \log_b N

.

3

Apply the power property to simplify the terms.

2\log_b 3 + \log_b 5 - 2

The power property states that

\log_b(M^k) = k\log_b M

. Also,

\log_b(b^2) = 2

because the base

b

raised to the second power is

b^2

.

4

Substitute the given values

x = \log_b 3

and

y = \log_b 5

.

2x + y - 2

Replacing the logarithmic expressions with

x

and

y

gives the final simplified expression.

Key Concept

Applying logarithmic properties (quotient, product, power) to simplify expressions

Estimated Time:1m 0s

Question 214Question →

What value of the base $b$ satisfies the equation $\log_b (3b + 10) = 2$ ?

Show answer & explanation

Answer: 5

Answer

The correct answer is 5.

Applying the definition of logarithms to the equation

\log_b(3b + 10) = 2

converts it to the quadratic equation

b^2 = 3b + 10

. Subtracting

3b

and

10

from both sides results in

b^2 - 3b - 10 = 0

. Factoring the quadratic yields

(b - 5)(b + 2) = 0

, giving the potential solutions

b = 5

and

b = -2

. Because the base of a logarithm must be positive (

b > 0

), the negative solution is extraneous. This leaves

5

as the only valid base.

Step-by-Step Solution

1

Convert the logarithm to exponential form.

b^2 = 3b + 10

By the definition of logarithms,

\log_b(x) = y

is equivalent to

b^y = x

.

2

Rearrange into standard quadratic form.

b^2 - 3b - 10 = 0

Subtracting

3b

and

10

from both sides sets the quadratic expression equal to zero.

3

Factor the quadratic equation.

(b - 5)(b + 2) = 0

Finding two integers that multiply to

-10

and add to

-3

gives

-5

and

2

.

4

Solve for the variable and apply base constraints.

b = 5

Solving the factored equation yields

b = 5

or

b = -2

. Since a logarithmic base must be strictly positive (

b > 0

), we reject

b = -2

as extraneous, leaving

b = 5

.

Key Concept

Converting logarithmic equations to exponential form and verifying base restrictions.

Estimated Time:1m 30s

Question 215Question →

The functions $f$ and $g$ are defined for all permissible real numbers by $f(x) = \frac{x + 3}{x - 1}$ and $g(x) = 2x - 5$ . If $(f \circ g)(x) = 3$ , what is the value of $x$ ?

Show answer & explanation

Answer: 4

Answer

The value of

x

that satisfies the equation

(f \circ g)(x) = 3

is

4

.

To solve for

x

in

(f \circ g)(x) = 3

, we find the composite function

f(g(x))

by substituting

g(x) = 2x - 5

into

f(x)

. This yields

f(g(x)) = \frac{(2x-5)+3}{(2x-5)-1} = \frac{2x-2}{2x-6}

. Setting this equal to

3

gives

\frac{2x-2}{2x-6} = 3

. Multiplying by

2x-6

yields

2x-2 = 6x-18

. Rearranging terms to isolate

x

gives

4x = 16

, which results in

x = 4

.

Step-by-Step Solution

1

Substitute the expression for

g(x)

into

f(x)

to obtain the composite function

(f \circ g)(x)

.

(f \circ g)(x) = \frac{2x - 2}{2x - 6}

By definition of function composition,

(f \circ g)(x) = f(g(x))

. Substituting

g(x) = 2x - 5

into

f(x) = \frac{x+3}{x-1}

yields

f(2x-5) = \frac{(2x-5)+3}{(2x-5)-1}

, which simplifies to

\frac{2x-2}{2x-6}

.

2

Set the composite function expression equal to

3

and solve for

x

.

x = 4

We set

\frac{2x-2}{2x-6} = 3

. Multiplying both sides by the denominator

2x-6

gives

2x-2 = 3(2x-6)

. Expanding the right side gives

2x-2 = 6x-18

. Subtracting

2x

from both sides and adding

18

to both sides results in

16 = 4x

. Dividing by

4

gives

x = 4

.

Key Concept

Function Composition and Evaluation

Estimated Time:1m 30s

Question 216Question →

Let the functions $f$ and $g$ be defined by $f(x) = |x - 4|$ and $g(x) = 2x + 1$ for all real numbers. For what values of $x$ does the composition $f(g(x)) = 5$ ?

x = -1

and

x = 4

x = 4

only

x = 2

and

x = 6

x = -2

and

x = 2

x = -4

and

x = 4

Show answer & explanation

Answer: $x = -1$ and $x = 4$

Answer

x = -1

and

x = 4

To find the values of

x

for which

f(g(x)) = 5

, we substitute the expression for

g(x)

into

f(x)

, yielding

|(2x + 1) - 4| = 5

, which simplifies to

|2x - 3| = 5

. Setting up the two possible cases for the absolute value gives

2x - 3 = 5

(which solves to

x = 4

) and

2x - 3 = -5

(which solves to

x = -1

). Both values of

x

satisfy the original composition equation.

Step-by-Step Solution

1

Substitute the inner function

g(x) = 2x + 1

into the outer function

f(x) = |x - 4|

to find the composite function

f(g(x))

.

f(g(x)) = |(2x + 1) - 4| = |2x - 3|

Function composition requires evaluating the outer function at the output of the inner function.

2

Set the composite expression equal to

5

.

|2x - 3| = 5

We are solving for the values of

x

where the output of the composite function is

5

.

3

Split the absolute value equation into its positive and negative cases and solve each linear equation for

x

.

Case 1:

2x - 3 = 5 \implies 2x = 8 \implies x = 4

. Case 2:

2x - 3 = -5 \implies 2x = -2 \implies x = -1

.

An absolute value equation

|u| = c

(with

c > 0

) is equivalent to

u = c

or

u = -c

.

Key Concept

Evaluating a composite function with an absolute value and solving the resulting equations by considering both positive and negative cases.

Estimated Time:1m 30s

Question 217Question →

A circle in the standard $(x, y)$ coordinate plane is defined by the equation $(x - 2)^2 + y^2 = 20$ , and a line is defined by the equation $y = 2x - 4$ . If the circle and the line intersect at two points, $(x_1, y_1)$ and $(x_2, y_2)$ , what is the value of $x_1 + x_2 + y_1 + y_2$ ?

Show answer & explanation

Answer: 4

Answer

The value of

x_1 + x_2 + y_1 + y_2

is 4.

Substituting the equation of the line into the circle's equation gives

(x-2)^2 + (2x-4)^2 = 20

. Since

(2x-4) = 2(x-2)

, we can write this as

(x-2)^2 + 4(x-2)^2 = 20

, which simplifies to

5(x-2)^2 = 20

, and further to

(x-2)^2 = 4

. Solving for

x

yields

x - 2 = 2

or

x - 2 = -2

, so

x = 4

or

x = 0

. Substituting these values into the linear equation gives the points

(4, 4)

and

(0, -4)

. The sum of all coordinates is

4 + 0 + 4 + (-4) = 4

.

Step-by-Step Solution

1

Substitute the linear equation into the circle equation.

(x - 2)^2 + (2x - 4)^2 = 20

To find the coordinates of the intersection points by reducing the system to a single variable equation.

2

Simplify the equation using algebraic substitution.

5(x - 2)^2 = 20

Factoring 2 out of the term

(2x-4)

yields

2(x-2)

. Squaring it results in

4(x-2)^2

. Combining this with the first term

(x-2)^2

simplifies the expression.

3

Solve for the two possible x-coordinates.

x = 0

and

x = 4

Dividing both sides by 5 yields

(x-2)^2 = 4

, which means

x-2 = \pm 2

.

4

Substitute the x-coordinates back into the linear equation to find the corresponding y-coordinates.

For

x = 0

,

y = -4

, giving the point

(0, -4)

. For

x = 4

,

y = 4

, giving the point

(4, 4)

.

To determine the coordinates of both intersection points.

5

Calculate the sum of all coordinates:

x_1 + x_2 + y_1 + y_2

.

0 + 4 + (-4) + 4 = 4

To find the final value requested by the question.

Key Concept

Solving systems of linear and quadratic (circular) equations by substitution and factoring.

Question 218Question →

If $x$ and $y$ are positive real numbers such that $\log(x + y) = \log x + \log y$ , which of the following equations expresses $y$ in terms of $x$ ?

y = \frac{x}{x+1}

y = \frac{x}{1-x}

y = \frac{x}{x-1}

(where

x > 1

)

y = x

Any positive real number

y

, because the equation is a general identity for all positive

x

and

y

Show answer & explanation

Answer: $y = \frac{x}{x-1}$ (where $x > 1$ )

Answer

y = \frac{x}{x-1}

(where

x > 1

)

By applying the product property of logarithms, the right side of the given equation

\log x + \log y

simplifies to

\log(xy)

. This transforms the equation to

\log(x+y) = \log(xy)

. Equating the arguments gives the algebraic equation

x+y=xy

. Rearranging this equation to group all

y

terms on one side yields

x = xy - y

. Factoring out

y

gives

x = y(x-1)

. Finally, dividing both sides by

x-1

results in the equation stating that

y

is equal to

x

divided by the quantity

x

minus one.

Step-by-Step Solution

1

Apply the product property of logarithms to the right side of the equation:

\log x + \log y = \log(xy)

.

The equation becomes

\log(x + y) = \log(xy)

.

This allows both sides to be expressed as a single logarithm with the same base.

2

Equate the arguments of the logarithms, since

\log(u) = \log(v)

implies

u = v

for positive real numbers.

x + y = xy

To eliminate the logarithms and work with a purely algebraic equation.

3

Isolate the terms containing

y

on one side of the equation.

x = xy - y

Grouping the

y

terms on the same side is a necessary step to factor out the variable we want to solve for.

4

Factor out

y

from the right side and divide both sides by the remaining coefficient

x - 1

.

x = y(x - 1) \implies y = \frac{x}{x-1}

(for

x > 1

to keep

y

positive and avoid division by zero).

This completely isolates

y

in terms of

x

.

Key Concept

Properties of Logarithms and Algebraic Isolation

Question 219Question →

If $8^{x - 2} = \left(\frac{1}{4}\right)^{x - 7}$ , what is the value of $x$ ?

Show answer & explanation

Answer: 4

Answer

The correct answer is

4

.

To solve

8^{x - 2} = \left(\frac{1}{4}\right)^{x - 7}

, express both bases as powers of

2

. Since

8 = 2^3

and

\frac{1}{4} = 2^{-2}

, the equation can be written as

(2^3)^{x-2} = (2^{-2})^{x-7}

. Using the power of a power property

(a^m)^n = a^{mn}

, we simplify this to

2^{3x - 6} = 2^{-2x + 14}

. Because the bases are equal, their exponents must be equal:

3x - 6 = -2x + 14

. Adding

2x

and

6

to both sides yields

5x = 20

, which simplifies to

x = 4

.

Step-by-Step Solution

1

Express both bases as powers of 2

8 = 2^3

and

\frac{1}{4} = 2^{-2}

To solve exponential equations with different bases, rewrite them using a common base.

2

Substitute the powers back into the equation and simplify exponents

2^{3(x-2)} = 2^{-2(x-7)}

Apply the exponent power rule

(a^m)^n = a^{mn}

.

3

Equate the exponents since the bases are identical

3(x-2) = -2(x-7)

If

b^y = b^z

and

b > 0, b \neq 1

, then

y = z

.

4

Solve the linear equation for

x

3x - 6 = -2x + 14 \implies 5x = 20 \implies x = 4

Distribute, group like terms, and isolate the variable.

Key Concept

Solving exponential equations using a common base

Question 220Question →

A software company's weekly revenue, $R$ , in thousands of dollars, is modeled by the function $R(x) = -x^2 + 8x$ , where $x$ represents the price of the software in dollars. The company's weekly operating cost, $C$ , in thousands of dollars, is modeled by the function $C(x) = 2x + 5$ . The company breaks even when its weekly revenue equals its weekly operating cost. What is the sum of the possible software prices, in dollars, at which the company breaks even?

Show answer & explanation

Answer: 6

Answer

The sum of the possible software prices at which the company breaks even is 6.

To find the software prices at which the company breaks even, we set the revenue function

R(x)

equal to the cost function

C(x)

, which gives

-x^2 + 8x = 2x + 5

. Rearranging this equation into standard quadratic form gives

x^2 - 6x + 5 = 0

. Factoring this expression yields

(x - 5)(x - 1) = 0

, meaning the individual solutions are

x = 5

and

x = 1

. The sum of these possible break-even software prices is

5 + 1 = 6

.

Step-by-Step Solution

1

Set the revenue function equal to the cost function.

-x^2 + 8x = 2x + 5

The break-even points occur where the revenue and cost are equal.

2

Rearrange the terms to set the equation to zero.

x^2 - 6x + 5 = 0

Moving all terms to one side of the equation puts it in standard quadratic form, allowing us to solve for

x

.

3

Factor the quadratic equation.

(x - 5)(x - 1) = 0

Factoring is a standard method to find the roots of a quadratic equation of the form

ax^2 + bx + c = 0

.

4

Find the individual roots by setting each factor to zero.

x = 5

and

x = 1

If the product of two factors is zero, at least one of the factors must be zero.

5

Calculate the sum of the solutions.

5 + 1 = 6

The question specifically asks for the sum of the possible prices.

Key Concept

Solving systems of linear and non-linear equations by setting the equations equal to each other and solving the resulting quadratic equation.

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