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SAT II Math I : Exponents and Logarithms

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Example Question #61 : Sat Subject Test In Math I

Give the set of real solutions to the equation

$3 ^{2x+1} -28 \cdot 3^{x}+9= 0$

(round to the nearest hundredth, if applicable)

Possible Answers:

$\left \{ -1, 2 \right \}$

$\left \{ 2 \right \}$

$\left \{ 1, 2 \right \}$

$\left \{ -1 \right \}$

The equation has no real solutions.

Correct answer:

$\left \{ -1, 2 \right \}$

Explanation:

Using the Product of Powers Rule, then the Power of a Power Rule, rewrite the first term:

$3 ^{2x+1} -28 \cdot 3^{x}+9= 0$

$3 ^{2x } \cdot 3^{1 } -28 \cdot 3^{x}+9= 0$

$3 \cdot 3 ^{2x } -28 \cdot 3^{x}+9= 0$

$3 \cdot (3 ^{x } )^{2 } -28 \cdot 3^{x}+9= 0$

Substitute for $3^{x}$ ; the equation becomes

$3t^{2 } -28t+9= 0$

which is quadratic in terms of . The trinomial might be factorable using the method, where we split the middle term with integers whose product is $3 \cdot 9 = 27$ and whose sum is . By trial and error, we find the integers to be and , so the equation can be rewritten as follows:

$3t^{2 } -27t -t +9= 0$

Factoring by grouping:

$(3t^{2 } -27t )-( t -9)= 0$

3t (t -9 )-1 ( t -9)= 0

(3t -1 )( t -9)= 0

By the Zero Product Rule, one of these two factors must be equal to 0.

If t - 9 = 0 , then t = 9 .

Since $3 ^{2} = 9$ , then substituting this as well as substituting $3^{x}$ back for , we get

$3^{x} = 3 ^{2}$ ,

and

x = 2

If 3t-1 = 0 , then

3t = 1

$t = \frac{1}{3}$

Since $3 ^{-1} = \frac{1}{3}$ , then substituting this as well as substituting $3^{x}$ back for , we get

$3^{x} = 3 ^{-1}$ , and

x= -1

The solution set is therefore $\left \{ -1, 2 \right \}$

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Example Question #11 : Exponents And Logarithms

Give the set of real solutions to the equation

$2 ^{2x+2} + 11 \cdot 2^{x} - 3 = 0$

(round to the nearest hundredth, if applicable)

Possible Answers:

$\left \{ -2, -0.58\right \}$

$\left \{ -0.58\right \}$

$\left \{ -2, 0.58\right \}$

$\left \{ 0.58\right \}$

$\left \{ -2 \right \}$

Correct answer:

$\left \{ -2 \right \}$

Explanation:

Using the Product of Powers Rule, then the Power of a Power Rule, rewrite the first term:

$2 ^{2x+2} + 11 \cdot 2^{x} - 3 = 0$

$2 ^{2x } \cdot 2 ^{2}+ 11 \cdot 2^{x} - 3 = 0$

$4 \cdot( 2 ^{x } ) ^{2}+ 11 \cdot 2^{x} - 3 = 0$

Substitute for $2^{x}$ ; the equation becomes

$4t ^{2}+ 11t- 3 = 0$ ,

which is quadratic in terms of . The trinomial might be factorable using the method, where we split the middle term with integers whose product is $4 \cdot (-3) = -12$ and whose sum is 11. By trial and error, we find the integers to be 12 and , so the equation can be written as follows:

$4t ^{2} -t + 12t- 3 = 0$

Factoring by grouping:

$(4t ^{2} -t )+( 12t- 3 )= 0$

t (4t -1 )+ 3 (4t -1 )= 0

(t + 3 )(4t -1 )= 0

By the Zero Product Rule, one of these two factors must be equal to 0.

If t+3 , then t = -3 .

Substituting $2^{x}$ back for , we get

$2^{x}= -3$ .

This is impossible, since any power of a positive number must be positive.

If 4t -1 = 0 , then:

4t =1

$t = \frac{1}{4}$

Substituting $2^{x}$ back for , we get

$2^{x}= \frac{1}{4}$

Since $\frac{1}{4} = \frac{1}{2^{2}} = 2^{-2}$ ,

it holds that $2^{x}= 2^{-2}$ , and x = -2 , the only solution.

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Example Question #11 : Exponents And Logarithms

Simplify $log_{10}(\frac{10}{100})$

Possible Answers:

Correct answer:

Explanation:

One of the properties of log is that $\small log_{x}(\frac{n}{m})=log_{x}(n)-log_{x}(m)$

Applying that principle to this problem:

$log_{10}(\frac{10}{100})=log_{10}(10)-log_{10}(100)$

Simplifying the log base 10

$log_{10}(10)=log_{10}(10^1)=1$

$log_{10}(100)=log_{10}(10^2)=2$

Plug in the values to the first equation:

$log_{10}(\frac{10}{100})=log_{10}(10)-log_{10}(100)=1-2=-1$

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Example Question #11 : Exponents And Logarithms

Evaluate: $xe^{ln(4-2x)}$

Possible Answers:

2x-4

(4x-2)^x

4x-2

-2x^2+4x

-2x^2-4x

Correct answer:

-2x^2+4x

Explanation:

An exponential base raised to the natural log will eliminate, leaving only the terms of the power. This is a log rule that can be used to simplify the expression.

$xe^{ln(4-2x)} = x(4-2x)$

Distribute the x variable through the binomial.

4x-2x^2

The answer is: -2x^2+4x

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Example Question #21 : Mathematical Relationships

$\textup{Solve for }x:$

$2 ^{2x} +4 = 2^{x+2}$

Possible Answers:

x = 2

$\textup{The equation has no solution}$

x = - 1

x = 1

x = 0

Correct answer:

x = 1

Explanation:

By the Power of a Power and Product of Power Rules, we can rewrite this equation as

$2 ^{2x} +4 = 2^{x+2}$

$(2 ^{ x }) ^{2} +4 = 2^{2} \cdot 2^{x}$

$(2 ^{ x }) ^{2} +4 = 4 \cdot 2^{x}$

Substitute for $2^{x}$ ; the resulting equation is the quadratic equation

$t^{2} +4 = 4t$ ,

which can be written in standard form by subtracting from both sides:

$t^{2} +4 - 4t = 4t - 4t$

$t^{2}- 4t +4 = 0$

The quadratic trinomial fits the perfect square trinomial pattern:

$t^{2}- 2 \cdot t \cdot 2 + 2^{2} = 0$

$(t-2)^{2} = 0$

By the square root principle,

t - 2= 0

t = 2

Substituting $2^{x}$ for :

$2^{x} = 2$

$2^{x} = 2 ^{1}$

x = 1

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Example Question #61 : Sat Subject Test In Math I

Solve for (round to the nearest hundredth):

$3 ^{x+ 4} = 2^{x+ 9}$

Possible Answers:

0.18

5.93

1.03

26.22

0.45

Correct answer:

0.45

Explanation:

Take the natural logarithm of both sides:

$3 ^{x+ 4} = 2^{x+ 9}$

$\ln 3 ^{x+ 4} = \ln 2^{x+ 9}$

By Logarithm of a Power Rule, the above becomes

$(x+ 4) \ln 3 = (x+ 9) \ln 2$

After distributing, solve for :

$x \ln 3 + 4 \ln 3 = x \ln 2+ 9 \ln 2$

$x \ln 3 + 4 \ln 3 - x \ln 2 - 4 \ln 3 = x \ln 2+ 9 \ln 2 - x \ln 2 - 4 \ln 3$

$x \ln 3 - x \ln 2 = 9 \ln 2 - 4 \ln 3$

Factor out the left side, then divide:

$x( \ln 3 - \ln 2) = 9 \ln 2 - 4 \ln 3$

$\frac{x( \ln 3 - \ln 2) }{\ln 3 - \ln 2}=\frac{ 9 \ln 2 - 4 \ln 3}{\ln 3 - \ln 2}$

$x=\frac{ 9 \ln 2 - 4 \ln 3}{\ln 3 - \ln 2}$

Substituting the values of the logarithms:

$x \approx \frac{ 9 (0.6931) - 4 (1.0986)}{1.0986 - 0.6931}$

$x\approx \frac{ 1.8439}{0.4055}$

$x \approx 0.45472$

This rounds to 0.45.

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Example Question #62 : Sat Subject Test In Math I

Solve for :

$2^{x} = 8 ^{2-x}$

Possible Answers:

$x = 1 \frac{1}{2}$

x = 3

x = 2

$x = \frac{1}{2}$

No solution

Correct answer:

$x = 1 \frac{1}{2}$

Explanation:

$8 = 2^{3}$ , so the equation

$2^{x} = 8 ^{2-x}$

can be rewritten as:

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

By the Power of a Power rule:

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

It follows that

x = 3 (2-x )

Solving for :

$x = 3 \cdot 2- 3 \cdot x$

x = 6 - 3x

x + 3x = 6 - 3x + 3x

4x = 6

$4x \div 4 = 6 \div 4$

$x = 1 \frac{1}{2}$

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SAT II Math I : Exponents and Logarithms

Study concepts, example questions & explanations for SAT II Math I

All SAT II Math I Resources

Example Questions

Example Question #61 : Sat Subject Test In Math I

Example Question #11 : Exponents And Logarithms

Example Question #11 : Exponents And Logarithms

Example Question #11 : Exponents And Logarithms

Example Question #21 : Mathematical Relationships

Example Question #61 : Sat Subject Test In Math I

Example Question #62 : Sat Subject Test In Math I

All SAT II Math I Resources

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