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SAT II Math II : Solving Exponential, Logarithmic, and Radical Functions

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

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Example Questions

Example Question #1 : Solving Exponential, Logarithmic, And Radical Functions

Rewrite as a single logarithmic expression:

$\ln (y+ 1) - \ln (y + 2) + \ln (y + 3)$ $\displaystyle \ln (y+ 1) - \ln (y + 2) + \ln (y + 3)$

Possible Answers:

$= \ln \frac{y^{2}+4y + 3}{y + 2}$ $\displaystyle = \ln \frac{y^{2}+4y + 3}{y + 2}$

$\ln \left (y + 4 \right )$ $\displaystyle \ln \left (y + 4 \right )$

$\ln \frac{y^{2}+3}{y + 2}$ $\displaystyle \ln \frac{y^{2}+3}{y + 2}$

$\ln \left (y + 2 \right )$ $\displaystyle \ln \left (y + 2 \right )$

$= \ln \frac{y+1}{y^{2}+5y+6}$ $\displaystyle = \ln \frac{y+1}{y^{2}+5y+6}$

Correct answer:

$= \ln \frac{y^{2}+4y + 3}{y + 2}$ $\displaystyle = \ln \frac{y^{2}+4y + 3}{y + 2}$

Explanation:

Using the properties of logarithms

$\ln a - \ln b = \ln \frac{a}{b}$ $\displaystyle \ln a - \ln b = \ln \frac{a}{b}$ and $\ln a + \ln b = \ln ab$ $\displaystyle \ln a + \ln b = \ln ab$,

simplify as follows:

$\ln (y+ 1) - \ln (y + 2) + \ln (y + 3)$ $\displaystyle \ln (y+ 1) - \ln (y + 2) + \ln (y + 3)$

$= \ln \frac{y+ 1}{y + 2} + \ln (y + 3)$ $\displaystyle = \ln \frac{y+ 1}{y + 2} + \ln (y + 3)$

$= \ln \left [\frac{y+ 1}{y + 2} \cdot (y + 3) \right ]$ $\displaystyle = \ln \left [\frac{y+ 1}{y + 2} \cdot (y + 3) \right ]$

$= \ln \frac{y^{2}+4y + 3}{y + 2}$ $\displaystyle = \ln \frac{y^{2}+4y + 3}{y + 2}$

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Example Question #1 : Solving Exponential, Logarithmic, And Radical Functions

Simplify by rationalizing the denominator:

$\frac{11}{6- \sqrt{3}}$ $\displaystyle \frac{11}{6- \sqrt{3}}$

Possible Answers:

$\frac{6- \sqrt{3}}{3 }$ $\displaystyle \frac{6- \sqrt{3}}{3 }$

$\frac{6+ \sqrt{3}}{3 }$ $\displaystyle \frac{6+ \sqrt{3}}{3 }$

$2- \sqrt{3}$ $\displaystyle 2- \sqrt{3}$

$2+ \sqrt{3}$ $\displaystyle 2+ \sqrt{3}$

$\frac{11\sqrt{3}}{3}$ $\displaystyle \frac{11\sqrt{3}}{3}$

Correct answer:

$\frac{6+ \sqrt{3}}{3 }$ $\displaystyle \frac{6+ \sqrt{3}}{3 }$

Explanation:

Multiply the numerator and the denominator by the conjugate of the denominator, which is $6 + \sqrt{3}$ $\displaystyle 6 + \sqrt{3}$. Then take advantage of the distributive properties and the difference of squares pattern:

$\frac{11}{6- \sqrt{3}}$ $\displaystyle \frac{11}{6- \sqrt{3}}$

$= \frac{11(6+ \sqrt{3})}{\left (6- \sqrt{3} \right )(6+ \sqrt{3})}$ $\displaystyle = \frac{11(6+ \sqrt{3})}{\left (6- \sqrt{3} \right )(6+ \sqrt{3})}$

$= \frac{11(6+ \sqrt{3})}{ 6^{2}- \left (\sqrt{3} \right ) ^{2}\right }$ $\displaystyle = \frac{11(6+ \sqrt{3})}{ 6^{2}- \left (\sqrt{3} \right ) ^{2}\right }$

$= \frac{11(6+ \sqrt{3})}{36-3 }$ $\displaystyle = \frac{11(6+ \sqrt{3})}{36-3 }$

$= \frac{11(6+ \sqrt{3})}{3 3 }$ $\displaystyle = \frac{11(6+ \sqrt{3})}{3 3 }$

$= \frac{6+ \sqrt{3}}{3 }$ $\displaystyle = \frac{6+ \sqrt{3}}{3 }$

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Example Question #3 : Solving Functions

Simplify:

$\sqrt[3]{\sqrt[2]{x^{-9}}}$ $\displaystyle \sqrt[3]{\sqrt[2]{x^{-9}}}$

You may assume that $\displaystyle x$ is a nonnegative real number.

Possible Answers:

$\frac{ \sqrt[3]{x}}{x }$ $\displaystyle \frac{ \sqrt[3]{x}}{x }$

$\frac{1}{x^{3}}$ $\displaystyle \frac{1}{x^{3}}$

$\frac{ \sqrt{x}}{x^{2}}$ $\displaystyle \frac{ \sqrt{x}}{x^{2}}$

$\sqrt[3]{x}$ $\displaystyle \sqrt[3]{x}$

$x \sqrt{x}$ $\displaystyle x \sqrt{x}$

Correct answer:

$\frac{ \sqrt{x}}{x^{2}}$ $\displaystyle \frac{ \sqrt{x}}{x^{2}}$

Explanation:

The best way to simplify a radical within a radical is to rewrite each root as a fractional exponent, then convert back.

First, rewrite the roots as exponents.

$\sqrt[3]{\sqrt[2]{x^{-9}}} = \sqrt[3]{\left ( x^{-9} \right ) ^{\frac{1}{2}} } = \left ( \left ( x^{-9} \right ) ^{\frac{1}{2}} \right )^{\frac{1}{3}} = x ^{-9 \cdot \frac{1}{2} \cdot \frac{1}{3}} } } = x ^{- \frac{3}{2} } }= \frac{1}{x^{\frac{3}{2}}}$ $\displaystyle \sqrt[3]{\sqrt[2]{x^{-9}}} = \sqrt[3]{\left ( x^{-9} \right ) ^{\frac{1}{2}} } = \left ( \left ( x^{-9} \right ) ^{\frac{1}{2}} \right )^{\frac{1}{3}} = x ^{-9 \cdot \frac{1}{2} \cdot \frac{1}{3}} } } = x ^{- \frac{3}{2} } }= \frac{1}{x^{\frac{3}{2}}}$

Then convert back to a radical and rationalizing the denominator:

$\frac{1}{x^{\frac{3}{2}}} = \frac{1}{\sqrt {x^{3}}} = \frac{1 \cdot \sqrt{x}}{\sqrt {x^{3} \cdot \sqrt{x}}} = \frac{ \sqrt{x}}{\sqrt {x^{4} }} =\frac{ \sqrt{x}}{x^{2}}$ $\displaystyle \frac{1}{x^{\frac{3}{2}}} = \frac{1}{\sqrt {x^{3}}} = \frac{1 \cdot \sqrt{x}}{\sqrt {x^{3} \cdot \sqrt{x}}} = \frac{ \sqrt{x}}{\sqrt {x^{4} }} =\frac{ \sqrt{x}}{x^{2}}$

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Example Question #1 : Solving Exponential, Logarithmic, And Radical Functions

Let $f(x) = 2^{x}+ 3^{2x}$ $\displaystyle f(x) = 2^{x}+ 3^{2x}$. What is the value of f(-2) $\displaystyle f(-2)$?

Possible Answers:

$\frac{85}{324}$ $\displaystyle \frac{85}{324}$

$\frac{1}{85}$ $\displaystyle \frac{1}{85}$

$\displaystyle -85$

$\textup{The answer is not given.}$ $\displaystyle \textup{The answer is not given.}$

$\displaystyle 85$

Correct answer:

$\frac{85}{324}$ $\displaystyle \frac{85}{324}$

Explanation:

Replace the integer as $\displaystyle x$.

$f(-2) = 2^{-2}+ 3^{2(-2)} = 2^{-2}+ 3^{-4}$ $\displaystyle f(-2) = 2^{-2}+ 3^{2(-2)} = 2^{-2}+ 3^{-4}$

Evaluate each negative exponent.

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

$3^{-4} = \frac{1}{3^4} = \frac{1}{81}$ $\displaystyle 3^{-4} = \frac{1}{3^4} = \frac{1}{81}$

Sum the fractions.

$\frac{1}{4}+\frac{1}{81} = \frac{1(81)}{4(81)}+\frac{1(4)}{81(4)} = \frac{85}{324}$ $\displaystyle \frac{1}{4}+\frac{1}{81} = \frac{1(81)}{4(81)}+\frac{1(4)}{81(4)} = \frac{85}{324}$

The answer is: $\frac{85}{324}$ $\displaystyle \frac{85}{324}$

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Example Question #1 : Solving Exponential, Logarithmic, And Radical Functions

Find $\displaystyle x$: $\sqrt{-3x-5} = 7$ $\displaystyle \sqrt{-3x-5} = 7$

Possible Answers:

$\displaystyle -54$

$\displaystyle -18$

$\displaystyle -4$

$\displaystyle -9$

$\displaystyle 18$

Correct answer:

$\displaystyle -18$

Explanation:

Square both sides to eliminate the radical.

$(\sqrt{-3x-5}) ^2= 7^2$ $\displaystyle (\sqrt{-3x-5}) ^2= 7^2$

$\displaystyle -3x-5 = 49$

Add five on both sides.

$\displaystyle -3x-5 +5= 49+5$

$\displaystyle -3x = 54$

Divide by negative three on both sides.

$\frac{-3x}{-3} = \frac{54}{-3}$ $\displaystyle \frac{-3x}{-3} = \frac{54}{-3}$

The answer is: $\displaystyle -18$

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SAT II Math II : Solving Exponential, Logarithmic, and Radical Functions

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

All SAT II Math II Resources

Example Questions

Example Question #1 : Solving Exponential, Logarithmic, And Radical Functions

Example Question #1 : Solving Exponential, Logarithmic, And Radical Functions

Example Question #3 : Solving Functions

Example Question #1 : Solving Exponential, Logarithmic, And Radical Functions

Example Question #1 : Solving Exponential, Logarithmic, And Radical Functions

All SAT II Math II Resources

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