Question 99·200 Super-Hard SAT Math Questions·Advanced Math

Type the given rule as $f(x)=\frac{5}{12}\cdot96^{(x+1)/5}$ in Desmos, then type $f(9)$. Desmos prints $3840$. Since $f(9)$ means put $9$ in for $x$, the target value is $k=3840$.

A zero exponent makes a nonzero base equal $1$. At $x=9$, $2^{x-9}=2^0=1$ and $3^{(x-9)/5}=3^0=1$, so a form with both of those powers shows its coefficient as $k$. Shift the exponents around the input whose output you want to show. By contrast, $2^{x-8}=2^1$ at $x=9$, so a coefficient beside it doesn't show $k$ by itself.

Turn on the Desmos graph and add $y=3840\cdot2^{x-9}\cdot3^{(x-9)/5}$. Its curve overlaps the original across the window; hide and show the new curve to check. An equivalent form must agree for every $x$, not only at $9$. Here both powers equal $1$ at $9$, so the coefficient visibly gives $k$. The answer is $f(x)=3840\cdot2^{x-9}\cdot3^{\frac{x-9}{5}}$. Choice D.

Since $96=2^5\cdot3$, replace the base in the given function:

A power of a product applies to each factor. Give both factors the outside exponent:

Multiply the exponents on the power of $2$:

To expose the value at $x=9$, separate each exponent into a fixed part and a part that becomes zero there:

For a sum of exponents, $a^{u+v}=a^ua^v$. Apply that rule to both powers: $f(x)=\frac{5}{12}2^{10}3^2\cdot2^{x-9}3^{\frac{x-9}{5}}$. The fixed factors can now become one coefficient.

Type $\frac{5}{12}2^{10}3^2$ in Desmos. It prints $3840$, so the exact rewrite is $f(x)=3840\cdot2^{x-9}\cdot3^{\frac{x-9}{5}}$. At $x=9$, both remaining powers equal $1$, making $k$ the coefficient. Choice D.