For the exponential function defined by , the value of is , where is a constant. Which choice gives an equivalent form of the function that shows the value of as the base?

Solution available with step-by-step explanation

Super-Hard SAT Math Question 199 | Free SAT Prep | Aniko

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Question 199·200 Super-Hard SAT Math Questions·Advanced Math

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For the exponential function $f$ defined by $f(x)=2^{4x-1}\cdot 8^{1-2x}$ , the value of $f\left(\frac{5}{8}\right)$ is $k$ , where $k$ is a constant. Which choice gives an equivalent form of the function $f$ that shows the value of $k$ as the base?

When a problem asks you to rewrite an exponential to “show” $k=f(a)$ as the base, first simplify the function as much as possible (often by rewriting everything in base 2 and combining exponents). Then compute $k$ exactly, write it as $k=2^p$ , solve $k^a=2$ to express 2 as a power of $k$ , and substitute to rewrite the entire function as a single power of $k$ .

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Combine into one power

Rewrite $8$ as $2^3$ , then use $2^a\cdot 2^b=2^{a+b}$ to simplify $f(x)$ to a single power of 2.

Evaluate at $x=\frac{5}{8}$

Once you have $f(x)=2^{\text{(something)}}$ , substitute $x=\frac{5}{8}$ to get $k$ as a power of 2, like $k=2^p$ .

Make the base $k$

If $k=2^p$ , find $a$ so that $k^a=2$ (match exponents). Then replace the 2 in $f(x)$ with $k^a$ and simplify.

Desmos Guide

Graph the original function

Enter $y_1=2^{4x-1}\cdot 8^{1-2x}$ .

Find $k$ from a table

In a table, set $x=5/8$ (or $0.625$ ) and read the corresponding $y_1$ value; this is $k$ .

Graph the candidate expression

Enter $y_2=(\sqrt[4]{8})^{\frac{8-8x}{3}}$ and check that $y_1$ and $y_2$ match (compare several table values or verify the graphs overlap).

Step-by-step Explanation

Rewrite $f(x)$ using a common base

Rewrite $8$ as $2^3$ and combine exponents:

\begin{aligned} f(x)&=2^{4x-1}\cdot 8^{1-2x} =2^{4x-1}\cdot (2^3)^{1-2x}\\ &=2^{4x-1}\cdot 2^{3(1-2x)} =2^{4x-1}\cdot 2^{3-6x}. \end{aligned}

Combine the powers of 2

Add exponents:

\begin{aligned} f(x)&=2^{(4x-1)+(3-6x)} =2^{2-2x}. \end{aligned}

Compute $k=f\left(\frac{5}{8}\right)$

Substitute $x=\frac{5}{8}$ into the simplified form:

\begin{aligned} k&=2^{2-2\left(\frac{5}{8}\right)} =2^{2-\frac{10}{8}} =2^{2-\frac{5}{4}} =2^{\frac{3}{4}}. \end{aligned}

Since $2^{3/4}=\sqrt[4]{2^3}=\sqrt[4]{8}$ , we have $k=\sqrt[4]{8}$ .

Rewrite 2 as a power of $k$

Because $k=2^{3/4}$ , raise both sides to the power $\frac{4}{3}$ :

k^{\frac{4}{3}}=\left(2^{\frac{3}{4}}\right)^{\frac{4}{3}}=2.

So $2=k^{4/3}$ .

Substitute and match to the choices

Rewrite

\begin{aligned} f(x)=2^{2-2x}&=(k^{\frac{4}{3}})^{2-2x} =k^{\frac{4}{3}(2-2x)} =k^{\frac{8-8x}{3}}. \end{aligned}

Since $k=\sqrt[4]{8}$ , this is

f(x)=(\sqrt[4]{8})^{\frac{8-8x}{3}},

which matches $f(x)=(\sqrt[4]{8})^{\frac{8-8x}{3}}$ .

Strategy

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Desmos Guide

Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation