Solving Exponential, Logarithmic, and Radical Functions

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SAT Subject Test in Math II › Solving Exponential, Logarithmic, and Radical Functions

1 - 5

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

CORRECT

Explanation

Square both sides to eliminate the radical.

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

Add five on both sides.

Divide by negative three on both sides.

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

The answer is:

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

$\frac{85}{324}$

CORRECT

$\frac{1}{85}$

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

Explanation

Replace the integer as .

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

Evaluate each negative exponent.

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

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

Sum the fractions.

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

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

Simplify:

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

You may assume that is a nonnegative real number.

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

CORRECT

$\sqrt[3]{x}$

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

$x \sqrt{x}$

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

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.

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

Then convert back to a radical and rationalizing the denominator:

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

Rewrite as a single logarithmic expression:

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

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

CORRECT

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

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

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

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

Explanation

Using the properties of logarithms

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

simplify as follows:

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

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

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

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

Simplify by rationalizing the denominator:

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

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

CORRECT

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

$2+ \sqrt{3}$

$2- \sqrt{3}$

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

Explanation

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