Exponential Functions and Equations Alıştırma Soruları — SAT — Sayfa 4

In the equation $\frac{25^x}{5^{x-3}} = 125^2$ , what is the value of $x$ ?

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Cevap: 3

Cevap

The correct answer is 3.

By writing all terms with a base of 5, the equation

\frac{25^x}{5^{x-3}} = 125^2

becomes

\frac{5^{2x}}{5^{x-3}} = 5^6

. Applying the quotient rule of exponents, the left side simplifies to

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

. Setting the exponents equal gives

x+3 = 6

, which yields

x = 3

.

Adım Adım Çözüm

1

Express all terms with a common base of 5.

25^x = (5^2)^x = 5^{2x}

and

125^2 = (5^3)^2 = 5^6

.

Expressing all exponential terms with the same base allows the exponents to be equated directly once simplified.

2

Substitute these expressions back into the original equation and simplify the left side using the quotient rule of exponents.

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

. The equation becomes

5^{x+3} = 5^6

.

The quotient rule states that dividing exponential terms with the same base requires subtracting the exponent of the denominator from the exponent of the numerator:

\frac{b^m}{b^n} = b^{m-n}

.

3

Equate the exponents and solve for

x

.

x + 3 = 6 \implies x = 3

.

Since the bases on both sides of the equation are equal, their exponents must also be equal: if

b^y = b^z

where

b > 0

and

b \neq 1

, then

y = z

.

Anahtar Kavram

Solving exponential equations by expressing terms with a common base and applying exponent rules.

Tahmini Süre:1m 30s

Soru 62Soru →

The table below shows some values of the exponential function $f$ , where $f(t) = p \cdot q^t$ for constants $p$ and $q$ .

$t$	$f(t)$
$0$	$80$
$2$	$180$
$4$	$405$

What is the value of $q$ ?

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Cevap: 1.5

Cevap

1.5

The correct answer is 1.5. Since the value of the function at

t = 0

is

80

, the initial value coefficient is

80

. At

t = 2

, the value is

180

, which gives the equation

80 \cdot q^2 = 180

. Solving for

q^2

yields

q^2 = 2.25

, and taking the positive square root gives

q = 1.5

.

Adım Adım Çözüm

1

Set up the general exponential equation using the initial value

f(0) = p \cdot q^0 = 80 \implies p = 80

The initial value at

t = 0

directly gives the coefficient

p

because

q^0 = 1

.

2

Substitute another point from the table to solve for the base

q

f(2) = 80 \cdot q^2 = 180 \implies q^2 = 2.25

Using the point

(2, 180)

allows us to write an equation with one variable,

q

.

3

Solve for

q

by taking the square root

q = 1.5

Since the base of an exponential function must be positive, we take the positive square root of 2.25.

Anahtar Kavram

Determining the base of an exponential function from a table of values

Soru 63Soru →

If $3^{2x - 1} = 81 \cdot 9^{y + 2}$ , which of the following equations correctly expresses $x$ in terms of $y$ ?

x = \frac{1}{2}y + \frac{7}{2}

x = y + \frac{7}{2}

x = y + \frac{9}{2}

x = 4y + \frac{17}{2}

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Cevap: $x = y + \frac{9}{2}$

Cevap

x = y + \frac{9}{2}

The correct equation is found by expressing

81

as

3^4

and

9^{y+2}

as

3^{2y+4}

. Applying the product rule for exponents, the right side becomes

3^{2y+8}

. Since the bases are the same, equating the exponents gives

2x - 1 = 2y + 8

. Solving for

x

yields

x = y + \frac{9}{2}

.

Adım Adım Çözüm

1

Rewrite all parts of the equation using a common base of

3

.

Since

81 = 3^4

and

9 = 3^2

, the term

9^{y+2}

becomes

(3^2)^{y+2} = 3^{2y+4}

. The equation can be rewritten as

3^{2x - 1} = 3^4 \cdot 3^{2y + 4}

.

Expressing all exponential terms with the same base allows the use of exponent rules to simplify the equation.

2

Simplify the product on the right side of the equation using the product rule for exponents,

a^m \cdot a^n = a^{m+n}

.

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

, which simplifies to

3^{2x - 1} = 3^{2y + 8}

.

Adding the exponents of terms with a common base simplifies the right side into a single exponential expression.

3

Set the exponents equal to each other.

2x - 1 = 2y + 8

If two exponential expressions with the same positive base (other than

1

) are equal, their exponents must be equal.

4

Solve for

x

in terms of

y

.

Add

1

to both sides to get

2x = 2y + 9

, then divide by

2

to obtain

x = y + \frac{9}{2}

.

This isolates the variable

x

to express it as a function of

y

.

Anahtar Kavram

Solving exponential equations by expressing terms with a common base and applying exponent laws.

Soru 64Soru →

If $9^{2x + 1} = \left(\frac{1}{27}\right)^{x - 2}$ , what is the value of $x$ ?

-8

1

\frac{4}{7}

-\frac{8}{7}

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Cevap: $\frac{4}{7}$

Cevap

\frac{4}{7}

The correct answer is

\frac{4}{7}

. By rewriting both sides of the equation with a common base of

3

, the equation becomes

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

. Applying the power of a power rule gives

3^{4x + 2} = 3^{-3x + 6}

. Since the bases are equal, the exponents must be equal, giving

4x + 2 = -3x + 6

. Solving for

x

results in

7x = 4

, which gives

x = \frac{4}{7}

.

Adım Adım Çözüm

1

Rewrite each side of the equation with a common base of

3

.

9 = 3^2

and

\frac{1}{27} = 3^{-3}

, so the equation becomes

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

.

Before solving an exponential equation, it is helpful to express the bases in terms of their common prime base.

2

Apply the exponent power rule

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

to simplify the exponents.

3^{2(2x + 1)} = 3^{-3(x - 2)} \implies 3^{4x + 2} = 3^{-3x + 6}

.

Simplifying the expressions on both sides allows for equating the exponents directly.

3

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

4x + 2 = -3x + 6 \implies 7x = 4 \implies x = \frac{4}{7}

.

Since the bases are equal, the powers can only be equal if their exponents are equal.

Anahtar Kavram

Solving exponential equations by expressing bases in terms of a common base and equating the exponents.

Tahmini Süre:1m 30s

Soru 65Soru →

The population of a species of fish in a lake can be modeled by the function $P(t) = P_0 \cdot 2^{\frac{t}{d}}$ , where $P_0$ is the initial population when the population was first measured, $t$ represents the time in years since it was first measured, and $d$ is a constant representing the doubling time in years. If the population of the fish doubles every 6 years, and the population after 18 years is 3,200, what was the initial population of the fish?

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Cevap: 400

Cevap

The initial population of the fish was 400.

By substituting the doubling time

d = 6

and the final population of 3,200 at

t = 18

into the exponential model

P(t) = P_0 \cdot 2^{\frac{t}{d}}

, we get

3,200 = P_0 \cdot 2^{\frac{18}{6}}

. Simplifying the exponent gives

3,200 = P_0 \cdot 2^3

, which simplifies further to

3,200 = 8P_0

. Dividing both sides by 8 yields

P_0 = 400

.

Adım Adım Çözüm

1

Identify the values for the known variables from the word problem.

d = 6

years and at

t = 18

years,

P(18) = 3,200

.

To substitute these values into the exponential growth function model.

2

Substitute the known values into the exponential function

P(t) = P_0 \cdot 2^{\frac{t}{d}}

.

3,200 = P_0 \cdot 2^{\frac{18}{6}}

To set up an equation to solve for the unknown parameter

P_0

.

3

Simplify the exponent and calculate the growth factor.

3,200 = P_0 \cdot 2^3 \Rightarrow 3,200 = 8P_0

Reducing the fractional exponent simplifies the equation.

4

Solve for the initial population

P_0

by dividing both sides of the equation by 8.

P_0 = 400

Isolating

P_0

gives the initial population of the fish.

Anahtar Kavram

Using an exponential function to model real-world growth and solving for the initial value.

Soru 66Soru →

If $8^{x + 1} = \left(\frac{1}{4}\right)^{2x - 3}$ , what is the value of $x$ ?

\frac{3}{7}

9

\frac{9}{7}

-\frac{9}{7}

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Cevap: $\frac{3}{7}$

Cevap

\frac{3}{7}

To solve the equation, we rewrite both bases using the common base

2

. Because

8 = 2^3

, the left side simplifies to

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

. Because

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

, the right side simplifies to

(2^{-2})^{2x - 3} = 2^{-4x + 6}

. Since the bases are now identical, their exponents must be equal:

3x + 3 = -4x + 6

. Solving this linear equation by adding

4x

to both sides gives

7x + 3 = 6

. Subtracting

3

from both sides gives

7x = 3

. Dividing by

7

yields the value of

\frac{3}{7}

.

Adım Adım Çözüm

1

Express both sides of the equation using a common base of

2

.

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

Since

8 = 2^3

and

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

, we can rewrite the terms with the same base to solve the exponential equation.

2

Apply the distributive property to simplify the exponents.

2^{3x + 3} = 2^{-4x + 6}

Multiplying the outer exponent by each term inside the parentheses simplifies the expression.

3

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

x

.

3x + 3 = -4x + 6 \implies 7x = 3 \implies x = \frac{3}{7}

When bases are equal and positive (and not 1), their exponents must be equal.

Anahtar Kavram

Solving exponential equations by expressing both sides with a common base and equating exponents.

Soru 67Soru →

The mass of a sample of a chemical compound in a reaction decays exponentially. The mass, in grams, of the sample $t$ hours after the reaction starts can be modeled by the function $M(t) = a \cdot b^t$ , where $a$ and $b$ are positive constants. If the mass of the sample is $18$ grams after $2$ hours and $8$ grams after $4$ hours, what is the initial mass, in grams, of the sample?

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Cevap: 40.5

Cevap

40.5

The initial mass of the sample is

40.5

grams (which can also be entered as the fraction

81/2

). This is found by setting up the two equations from the given points:

a \cdot b^2 = 18

and

a \cdot b^4 = 8

. Dividing the second equation by the first eliminates

a

and gives

b^2 = \frac{4}{9}

. Substituting

b^2 = \frac{4}{9}

back into the first equation yields

a \cdot \frac{4}{9} = 18

. Multiplying both sides by

\frac{9}{4}

results in

a = 40.5

. Since

M(0) = a \cdot b^0 = a

, the initial mass of the sample is

40.5

grams.

Adım Adım Çözüm

1

Set up the system of exponential equations using the given coordinates.

a \cdot b^2 = 18

and

a \cdot b^4 = 8

This represents the mass of the sample at

t = 2

and

t = 4

using the model

M(t) = a \cdot b^t

.

2

Divide the second equation by the first equation to eliminate the constant

a

and solve for

b^2

.

b^2 = \frac{4}{9}

Dividing the equations yields

\frac{a \cdot b^4}{a \cdot b^2} = \frac{8}{18}

, which simplifies to

b^2 = \frac{4}{9}

.

3

Substitute the value of

b^2

back into the first equation to solve for the initial mass

a

.

a = 40.5

Substituting

b^2

gives

a \cdot \frac{4}{9} = 18

. Multiplying both sides by

\frac{9}{4}

yields

a = 18 \cdot \frac{9}{4} = 40.5

.

Anahtar Kavram

Solving systems of exponential equations to determine the initial value and decay factor.

Soru 68Soru →

If $4^{a - 1} \cdot 8^{2a} = 16^3$ , what is the value of $a$ ?

\\frac{9}{8}

\\frac{13}{8}

\\frac{7}{4}

\\frac{11}{4}

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Cevap: $\\frac{7}{4}$

Cevap

\\frac{7}{4}

The correct answer is

\frac{7}{4}

. Rewriting each base in the equation

4^{a - 1} \cdot 8^{2a} = 16^3

as a power of 2 gives

(2^2)^{a-1} \cdot (2^3)^{2a} = (2^4)^3

. Applying the power of a power rule results in

2^{2a-2} \cdot 2^{6a} = 2^{12}

. Using the product rule of exponents to combine the left side yields

2^{2a-2+6a} = 2^{8a-2} = 2^{12}

. Setting the exponents equal gives

8a - 2 = 12

, which simplifies to

8a = 14

, or

a = \frac{7}{4}

.

Adım Adım Çözüm

1

Rewrite each base in the equation

4^{a - 1} \cdot 8^{2a} = 16^3

as a power of 2.

(2^2)^{a-1} \cdot (2^3)^{2a} = (2^4)^3

To solve an exponential equation with different bases, rewrite the bases so they are identical.

2

Apply the power of a power rule

(x^m)^n = x^{m \cdot n}

to simplify each term.

2^{2a-2} \cdot 2^{6a} = 2^{12}

This simplifies the exponents by multiplying the inner and outer exponents.

3

Apply the product rule

x^m \cdot x^n = x^{m+n}

to combine the terms on the left side.

2^{8a-2} = 2^{12}

This combines the exponents of the terms with the common base of 2.

4

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

a

.

8a - 2 = 12 \Rightarrow 8a = 14 \Rightarrow a = \\frac{7}{4}

Since the bases are equal, their exponents must be equal.

Anahtar Kavram

Solving exponential equations by expressing all terms with a common base and applying exponent rules.

Alternatif Yöntem

Instead of converting to base 2, all terms can be written in base 4:

4^{a - 1} \cdot (4^{1.5})^{2a} = (4^2)^3

, which simplifies to

4^{a - 1 + 3a} = 4^6

, leading to

4a - 1 = 6

and

a = \\frac{7}{4}

.

Tahmini Süre:1m 30s

Soru 69Soru →

If $27^{x-1} = \sqrt{3^{5x+1}}$ , what is the value of $x$ ?

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Cevap: 7

Cevap

The correct answer is 7.

The correct answer is 7. By converting the base of 27 to

3^3

and rewriting the radical on the right side as a fractional exponent of

\frac{1}{2}

, the equation becomes

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

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

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

. Multiplying both sides by 2 yields

6x - 6 = 5x + 1

. Subtracting

5x

and adding 6 to both sides isolates

x

, giving the solution

x = 7

.

Adım Adım Çözüm

1

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

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

Expressing terms with a common base allows the exponents to be compared directly.

2

Apply the power of a power exponent rule,

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

, to simplify the exponents on both sides.

3^{3x-3} = 3^{\frac{5x+1}{2}}

Simplifying the expressions makes it possible to set the exponent expressions equal to each other.

3

Equate the exponents since the bases are identical.

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

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

4

Solve the linear equation for

x

by clearing the fraction and isolating the variable.

x = 7

Multiplying both sides by 2 gives

6x - 6 = 5x + 1

. Subtracting

5x

from both sides and adding 6 to both sides isolates

x

, resulting in

x = 7

.

Anahtar Kavram

Solving exponential equations by converting to a common base and applying exponent laws.

Exponential Functions and Equations

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