Question

rewrite using a single positive exponent.
$(4^2)^{-8}$

Explanation:

Step1: Recall exponent rule $(a^m)^n = a^{m\times n}$

For $(4^2)^{-8}$, apply the rule: $m = 2$, $n = -8$, so $4^{2\times(-8)} = 4^{-16}$

Step2: Recall negative exponent rule $a^{-n}=\frac{1}{a^n}$, but we need positive exponent. Wait, no—wait, actually, to rewrite with positive exponent, we can also use the property that $(a^m)^n = a^{mn}$, and then if we have a negative exponent, we can take reciprocal, but wait, the problem says "single positive exponent"—wait, no, maybe I made a mistake. Wait, no: actually, when we have $(a^m)^n = a^{m\times n}$, so $(4^2)^{-8} = 4^{2\times(-8)} = 4^{-16}$. But to write with positive exponent, we use $a^{-n} = \frac{1}{a^n}$, but that would be $\frac{1}{4^{16}}$, but that's a fraction. Wait, no—wait, maybe the problem allows expressing as a positive exponent in the denominator? Wait, no, the question says "rewrite using a single positive exponent"—wait, maybe I misread. Wait, no, the original expression is $(4^2)^{-8}$. Let's re-express:

Wait, actually, the exponent rule for power of a power is $(a^m)^n = a^{m \times n}$. So $(4^2)^{-8} = 4^{2 \times (-8)} = 4^{-16}$. But a negative exponent means reciprocal with positive exponent, so $4^{-16} = \frac{1}{4^{16}}$. But $\frac{1}{4^{16}}$ has a positive exponent in the denominator. Alternatively, maybe the problem expects using the rule that $(a^m)^{-n} = (a^{-m})^n = \frac{1}{(a^m)^n}$, but to write with a single positive exponent, we can write it as $\frac{1}{4^{16}}$, but that's a fraction. Wait, maybe I made a mistake in the sign. Wait, no: $(4^2)^{-8} = 4^{2 \times (-8)} = 4^{-16} = \frac{1}{4^{16}}$. So the single positive exponent is in the denominator, $4^{16}$, and the expression is $\frac{1}{4^{16}}$, but if we consider that the problem allows the positive exponent in the denominator, then that's the form. Alternatively, maybe the problem has a typo, but following the rules:

First, apply power of a power: $(4^2)^{-8} = 4^{2 \times (-8)} = 4^{-16}$

Then, convert negative exponent to positive by reciprocal: $4^{-16} = \frac{1}{4^{16}}$

So the expression with a single positive exponent (in the denominator) is $\frac{1}{4^{16}}$, but if we consider that the base can be inverted, but no—wait, maybe the problem actually wants to use the rule that $(a^m)^{-n} = (a^{-m})^n$, but that's still negative. Wait, no, the key is that to have a single positive exponent, we can write it as $\frac{1}{4^{16}}$, where $16$ is positive.

Wait, but maybe I messed up the exponent rule. Let's recheck: $(a^m)^n = a^{m \times n}$. So $(4^2)^{-8} = 4^{2 \times (-8)} = 4^{-16}$. Then, using $a^{-n} = \frac{1}{a^n}$, so $4^{-16} = \frac{1}{4^{16}}$. So the expression with a single positive exponent is $\frac{1}{4^{16}}$, or equivalently, $4^{-16}$ is not positive, so we need to make the exponent positive by taking reciprocal, resulting in $\frac{1}{4^{16}}$, which has a positive exponent of $16$ in the denominator.

Answer:

$\frac{1}{4^{16}}$ (or equivalently, if we consider the positive exponent in the base's reciprocal, but the standard way is $\frac{1}{4^{16}}$ with positive exponent $16$)