In the given equations, and are constants, and . Which choice is the value of ?

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Super-Hard SAT Math Question 1 | Free SAT Prep | Aniko

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Question 1·200 Super-Hard SAT Math Questions·Advanced Math

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In the given equations, $a$ and $b$ are constants, $a>1$ and $b>1$ . Which choice is the value of $x$ ?

\sqrt[8]{a^3}=\sqrt[6]{b^5}

a^{2x-1}=b^5

For systems like this, convert every radical into a rational exponent and eliminate one variable. A fast path is: (1) rewrite radicals as powers, (2) raise both sides to clear denominators, (3) solve for one variable as a power of the other, and (4) substitute so both sides share the same base; with $a>1$ , you can safely set the exponents equal.

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Convert radicals to exponents

Rewrite $\sqrt[8]{a^3}$ and $\sqrt[6]{b^5}$ using fractional exponents.

Clear the fractions in the exponents

After rewriting, raise both sides to a power that makes the exponents integers (use the LCM of the denominators).

Get one base

Use the first equation to write $b$ as a power of $a$ , then substitute into the second equation so both sides are powers of $a$ .

Desmos Guide

Choose a valid value for $a$ and compute $b$ from the first equation

Set a slider, for example a=2. Rewrite the first equation as $b=a^{9/20}$ , then enter b=a^(9/20) so Desmos computes $b$ from your chosen $a$ .

Graph both sides of the second equation as functions of $x$

Enter y1=a^(2x-1) and y2=b^5. Here y2 will be a horizontal line because $b^5$ is a constant once $b$ is defined.

Find the intersection x-value

Click the intersection point of $y_1$ and $y_2$ and read the $x$ -coordinate. (You can adjust the window or zoom if needed.) That $x$ -value is the solution.

Step-by-step Explanation

Rewrite the radical equation using exponents

\sqrt[8]{a^3}=\sqrt[6]{b^5} \quad\Longrightarrow\quad a^{3/8}=b^{5/6}

Raise both sides to the 24th power (the LCM of 8 and 6) to clear denominators:

a^{(3/8)\cdot 24}=b^{(5/6)\cdot 24} \quad\Longrightarrow\quad a^9=b^{20}

Express $b$ as a power of $a$

From $a^9=b^{20}$ and $a>1, b>1$ , take the $20$ th root:

b=a^{9/20}

Substitute into the second equation and solve for $x$

Substitute $b=a^{9/20}$ into $a^{2x-1}=b^5$ :

\begin{aligned} a^{2x-1} &= \left(a^{9/20}\right)^5 \\ &= a^{45/20} \\ &= a^{9/4} \end{aligned}

Since $a>1$ , equal powers of $a$ must have equal exponents:

2x-1=\frac{9}{4}

2x=\frac{13}{4} \quad\Longrightarrow\quad x=\frac{13}{8}

So the correct choice is $\frac{13}{8}$ .

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Step-by-step Explanation

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Strategy

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Desmos Guide

Step-by-step Explanation