Logarithmic and Exponential Expressions and Equations Alıştırma Soruları — ACT — Sayfa 2

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

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Cevap: 3

Cevap

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.

Adım Adım Çözüm

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.

Anahtar Kavram

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

Soru 22Soru →

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

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Cevap: 7

Cevap

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.

Adım Adım Çözüm

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

.

Anahtar Kavram

Solving logarithmic equations using properties of logarithms

Tahmini Süre:1m 30s

Soru 23Soru →

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

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Cevap: $2x + y - 2$

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

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

Tahmini Süre:1m 0s

Soru 24Soru →

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

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Cevap: 5

Cevap

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.

Adım Adım Çözüm

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

.

Anahtar Kavram

Converting logarithmic equations to exponential form and verifying base restrictions.

Tahmini Süre:1m 30s

Soru 25Soru →

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

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

Cevap

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.

Adım Adım Çözüm

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

.

Anahtar Kavram

Properties of Logarithms and Algebraic Isolation

Soru 26Soru →

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

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Cevap: 4

Cevap

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

.

Adım Adım Çözüm

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.

Anahtar Kavram

Solving exponential equations using a common base

Soru 27Soru →

If $x$ is a real number such that $\log_5(x) + \log_5(x - 20) = 3$ , what is the value of $x$ ?

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Cevap: 25

Cevap

The value of

x

is 25.

Applying the logarithmic product rule simplifies the equation to

\log_5(x^2 - 20x) = 3

. Writing this in exponential form yields

x^2 - 20x = 125

. Rearranging into standard form gives

x^2 - 20x - 125 = 0

, which factors into

(x - 25)(x + 5) = 0

. This gives potential solutions of

25

and

-5

. Because the logarithmic arguments must be strictly positive,

x = -5

is extraneous. Therefore, the only correct value is 25.

Adım Adım Çözüm

1

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

\log_5(x(x - 20)) = 3

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

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

.

2

Rewrite the logarithmic equation in exponential form.

x(x - 20) = 5^3 \implies x^2 - 20x = 125

The logarithmic equation

\log_b(y) = c

is equivalent to the exponential equation

b^c = y

.

3

Rearrange the quadratic equation into standard form and solve by factoring.

x^2 - 20x - 125 = 0 \implies (x - 25)(x + 5) = 0 \implies x = 25 \text{ or } x = -5

Subtracting 125 from both sides sets the quadratic equation to 0, which can then be factored into binomials whose product is 0.

4

Verify the potential solutions in the original equation to identify any extraneous roots.

For

x = -5

, the arguments of the original logarithms are negative, which is undefined. For

x = 25

, the arguments are positive. Thus, the only valid solution is

x = 25

.

Logarithmic functions are only defined for positive real numbers. Therefore, we must have

x > 0

and

x - 20 > 0

, which requires

x > 20

.

Anahtar Kavram

Solving logarithmic equations by combining logarithmic terms and checking for extraneous solutions.

Tahmini Süre:1m 30s

Soru 28Soru →

If $9^{2x - 1} = 27^{x + 4}$ , what is the value of $x$ ?

5

-13

14

6

-4

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Cevap: 14

Cevap

14

Converting both bases to 3 yields

(3^2)^{2x - 1} = (3^3)^{x + 4}

. Applying the exponent power rule gives

3^{4x - 2} = 3^{3x + 12}

. Equating the exponents results in

4x - 2 = 3x + 12

. Subtracting

3x

from both sides gives

x - 2 = 12

, and adding 2 to both sides results in

x = 14

. This matches the correct value of 14.

Adım Adım Çözüm

1

Express both sides of the equation with a common base of 3.

(3^2)^{2x - 1} = (3^3)^{x + 4}

Since

9 = 3^2

and

27 = 3^3

, rewriting the bases allows us to equate the exponents later.

2

Apply the power of a power property,

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

, to simplify the exponents on both sides.

3^{2(2x - 1)} = 3^{3(x + 4)}

which simplifies to

3^{4x - 2} = 3^{3x + 12}

To simplify an exponent raised to another power, multiply the exponents, ensuring the multiplier is distributed to both terms inside each exponent expression.

3

Set the exponents equal to each other and solve the resulting linear equation for

x

.

4x - 2 = 3x + 12 \implies x = 14

If two exponential expressions with the same positive base (other than 1) are equal, their exponents must be equal.

Anahtar Kavram

Solving exponential equations by expressing bases with a common base and equating the exponents.

Tahmini Süre:1m 30s

Logarithmic and Exponential Expressions and Equations

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