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SAT II Math II : Exponents and Logarithms

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

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Example Question #1 : Exponents And Logarithms

Solve for :

$5^{x} = 432$

Give the solution to the nearest hundredth.

Possible Answers:

$x \approx 2.86$

$x \approx 3.77$

$x\approx 3.37$

$x \approx 2.44$

$x \approx 1.94$

Correct answer:

$x \approx 3.77$

Explanation:

One way is to take the common logarithm of both sides and solve:

$5^{x} = 432$

$\log \left (5^{x} \right )= \log 432$

$x \log 5= \log 432$

$x = \frac{\log 432}{\log 5} \approx \frac{2.6355}{0.6990} \approx 3.77$

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Example Question #2 : Exponents And Logarithms

Solve for :

$\left (e+1 \right ) ^{k} = 100$

Give your answer to the nearest hundredth.

Possible Answers:

$k \approx 1.69$

$k \approx 3.61$

$k \approx 1.43$

$k \approx 3.51$ '

$k \approx 1.89$

Correct answer:

$k \approx 3.51$ '

Explanation:

Take the common logarithm of both sides and solve for :

$\left (e+1 \right ) ^{k} = 100$

$\log \left (e+1 \right ) ^{k} = \log 100$

$k \log \left (e+1 \right ) = 2$

$k= \frac{2}{ \log \left (e+1 \right ) }$

$k= \frac{2}{ \log \left (2.7183+1 \right ) } \approx \frac{2}{ \log 3.7183 } \approx \frac{2}{1.3133} \approx 3.51$

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Example Question #1 : Exponents And Logarithms

Solve for :

$10^{t} + 1 = e$

Give your answer to the nearest hundredth.

Possible Answers:

$t \approx -0.57$

$t \approx 0.42$

The equation has no solution.

$t \approx 0.46$

$t \approx 0.24$

Correct answer:

$t \approx 0.24$

Explanation:

$10^{t} + 1 = e$

$10^{t} = e - 1$

Take the common logarithm of both sides and solve for :

$\log 10^{t} = \log \left (e - 1 \right )$

$t = \log (e - 1) \approx \log (2.718-1) \approx \log 1.718 \approx 0.24$

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Example Question #1 : Exponents And Logarithms

Solve for :

$e^{x} = \pi$

Give your answer to the nearest hundredth.

Possible Answers:

$x \approx 0.87$

$x \approx 1.14$

$x \approx 2.01$

The equation has no solution.

$x \approx 0.50$

Correct answer:

$x \approx 1.14$

Explanation:

Take the natural logarithm of both sides and solve for :

$e^{x} = \pi$

$\ln e^{x} = \ln \pi$

$x = \ln \pi \approx \ln 3.1416 \approx 1.14$

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Example Question #5 : Exponents And Logarithms

To the nearest hundredth, solve for :

$2^{x+4} = 5^{x}$

Possible Answers:

$x \approx 6.32$

$x \approx 3.03$

$x \approx 1.20$

$x \approx 4.48$

$x \approx -1.68$

Correct answer:

$x \approx 3.03$

Explanation:

Take the common logarithm of both sides, then solve the resulting linear equation.

$2^{x+4} = 5^{x}$

$\log 2^{x+4} = \log 5^{x}$

$(x + 4) \log 2 = x \log 5$

$x\log 2 + 4\log 2 = x \log 5$

$4\log 2 = x \log 5 - x\log 2$

$4\log 2 = x\left ( \log 5 - \log 2 \right )$

$x = \frac{4\log 2 }{ \log 5 - \log 2 }$

$x \approx \frac{4 \cdot 0.3010}{0.6990-0.3010} \approx 3.03$

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Example Question #6 : Exponents And Logarithms

To the nearest hundredth, solve for :

$2^{x+10}= 10^{x+2}$

Possible Answers:

$x\approx 7.17$

The equation has no solution.

$x \approx 3.85$

$x \approx 2.31$

$x \approx 1.45$

Correct answer:

$x \approx 1.45$

Explanation:

Take the common logarithm of both sides, then solve the resulting linear equation.

$2^{x+10}= 10^{x+2}$

$\log 2^{x+10}= \log 10^{x+2}$

$\left (x+10 \right )\log 2 = x+2$

$x\log 2 +10 \log 2 = x+2$

$x\log 2 +10 \log 2 - 2 - x \log 2= x+2 - 2 - x \log 2$

$-2 +10 \log 2 = x - x \log 2$

$x \left (1 - \log 2 \right )= -2 +10 \log 2$

$x =\frac{ -2 +10 \log 2}{1 - \log 2 }$

$x \approx \frac{ -2 +10 \cdot 0.3010}{1 - 0.3010 } \approx 1.45$

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Example Question #1 : Exponents And Logarithms

Solve for :

$100 ^{x}= 66$

Possible Answers:

$x = \frac{ \log 66}{2}$

$x = \log 33$

$x = \log 64$

$x = 2- \log 66$

$x = \frac{2}{\log 66}$

Correct answer:

$x = \frac{ \log 66}{2}$

Explanation:

Take the common logarithm of both sides:

$100 ^{x}= 66$

$\log 100 ^{x}= \log 66$

$x \log 100 = \log 66$

$x \cdot 2 = \log 66$

$x = \frac{ \log 66}{2}$

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Example Question #8 : Exponents And Logarithms

Solve for :

$3 + \log 4 + \log 6 = \log N$

Possible Answers:

N = 24,000

N = 13,824

N = 1,000

N = 10,000

N = 72

Correct answer:

N = 24,000

Explanation:

The base of the common logarithm is 10, so

$3 = \log 10^{3} = \log 1,000$

The sum of three logarithms is the logarithm of the product of the three powers, so:

$3 + \log 4 + \log 6$

$=\log 1,000 + \log 4 + \log 6$

$= \log (1,000 \times 4 \times 6) = \log 24,000$

Therefore, N = 24,000

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Example Question #9 : Exponents And Logarithms

To the nearest hundredth, solve for :

$5^{N} = 10^{4-N}$

Possible Answers:

$N \approx 1.30$

$N \approx 13.29$

$N \approx 1.53$

$N \approx 2.35$

$N \approx 0.37$

Correct answer:

$N \approx 2.35$

Explanation:

$5^{N} = 10^{4-N}$

$\log 5^{N} = \log 10^{4-N}$

$N \log 5 = 4 - N$

$N \log 5 + N = 4$

$N \left (1+ \log 5 \right ) = 4$

$N = \frac{4}{1+ \log 5 } \approx \frac{4}{1+ 0.6990 } \approx 2.35$

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Example Question #10 : Exponents And Logarithms

Solve for :

$\log 20 - 2 + \log 6 = \log N$

Possible Answers:

$N = \frac{60}{e}$

The equation has no solution

$N = \frac{120}{e^{2}}$

N = 60

$N = 1 \frac{1}{5}$

Correct answer:

$N = 1 \frac{1}{5}$

Explanation:

The base of the common logarithm is 10, so

$2 = \log 10^{2} = \log 1 00$

The sum of logarithms is the logarithm of the product of the three powers, and the difference of logarithms is the logarithm of the quotient of their powers. Therefore,

$\log 20 - 2 + \log 6$

$= \log 20 - \log 100 + \log 6$

$= \log \frac{20 \times 6 }{100} = \log \frac{120 }{100} =\log \frac{6}{5}$

$N = \frac{6}{5} = 1 \frac{1}{5}$

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SAT II Math II : Exponents and Logarithms

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

All SAT II Math II Resources

Example Questions

Example Question #1 : Exponents And Logarithms

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Example Question #1 : Exponents And Logarithms

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