Use $(a^{-1})^m=a^{-m}$ (or $a^{-m}=\frac{1}{a^m}$) to simplify before deciding whether the function increases on $x\ge0$.

Once you have $2^{5x}$, use $a^{kx}=(a^k)^x$ to rewrite it as $(2^5)^x=(32)^x. Ѕ​οu​rce‍:⁠ anіko.аi

In Desmos, enter

• f(x)=(1/8)^(4-(5/3)x)

Add a restriction to view the relevant domain by graphing y=f(x) {x>=0}.

Use the table feature: create a table of $x$ values (for example $0,1,2$) and observe that $f(x)$ increases as $x$ increases. This supports that the minimum on $x\ge 0$ occurs at $x=0$. W​rit⁠t⁠en by ‌Аn​іk​о

Evaluate $f(0)$ in the table (or type f(0)) to get the minimum value $c$. Then check which answer choice has that coefficient in front of $(32)^x$.

Rewrite the base and simplify the exponent:

Since $8=2^3$,

In $f(x)=2^{5x-12}$, the base $2>1$ and the exponent $5x-12$ increases as $x$ increases.

So $f(x)$ is increasing on $x\ge 0$, meaning the minimum value occurs at the left endpoint $x=0$. Therefore, $c=f(0)$.

Separate the constant part from the $x$-dependent part: Content​ by Anік⁠о.а‌i

Because the minimum occurs at $x=0$, we have $c=f(0)=2^{-12}=\frac{1}{2^{12}}=\frac{1}{4{,}096}$.

So