Question

simplify the expression using exponent rules. leave your answer in exponential form with a positive exponent. (1 point)\\((2^{-5} \times 2^{2})^{4} = \square\\)

Explanation:

Step1: Apply the product rule of exponents ($a^m \times a^n = a^{m + n}$) inside the parentheses.

For the base 2, we have $-5 + 2 = -3$, so the expression inside the parentheses becomes $2^{-3}$.

Step2: Apply the power of a power rule of exponents ($(a^m)^n = a^{m \times n}$).

Now we raise $2^{-3}$ to the power of 4. So we multiply the exponents: $-3 \times 4 = -12$, giving $2^{-12}$.

Step3: Convert the negative exponent to a positive exponent using the rule $a^{-n} = \frac{1}{a^n}$, but we can also rewrite it as $2^{-12} = \frac{1}{2^{12}}$, however, the problem says to leave it in exponential form with a positive exponent. Wait, actually, we can also use the rule that $a^{-n}=\frac{1}{a^n}$, but if we want to express it with a positive exponent in the numerator, we can note that $2^{-12} = \frac{1}{2^{12}}$, but maybe we made a mistake. Wait, no, let's check again. Wait, step 1: $2^{-5} \times 2^{2} = 2^{-5 + 2} = 2^{-3}$. Then step 2: $(2^{-3})^4 = 2^{-3 \times 4} = 2^{-12}$. But to write it with a positive exponent, we use $a^{-n}=\frac{1}{a^n}$, so $2^{-12}=\frac{1}{2^{12}}$, but the problem says "exponential form with a positive exponent". Wait, maybe I misread. Wait, the problem says "Leave your answer in exponential form with a positive exponent". So actually, $2^{-12}$ can be written as $\frac{1}{2^{12}}$, but that's a fraction. Wait, no, maybe the question allows the base with a positive exponent, so $2^{-12} = \frac{1}{2^{12}}$, but let's check the exponent rules again. Wait, no, the product rule: when multiplying exponents with the same base, add the exponents. Then power of a power: multiply the exponents. So $(2^{-5} \times 2^{2})^4 = (2^{-5 + 2})^4 = (2^{-3})^4 = 2^{-12} = \frac{1}{2^{12}}$. But maybe the problem considers $2^{-12}$ as exponential form, but with a negative exponent. Wait, the problem says "positive exponent". So we need to rewrite $2^{-12}$ as $\frac{1}{2^{12}}$, but that's a fraction. Wait, no, maybe I made a mistake in the exponent addition. Wait, $-5 + 2 = -3$, correct. Then $-3 \times 4 = -12$, correct. So $2^{-12} = \frac{1}{2^{12}}$. But let's check with another approach. Alternatively, maybe the problem accepts $2^{-12}$ but with a positive exponent, so we can write it as $\frac{1}{2^{12}}$, but that's a fraction. Wait, no, maybe the question has a typo, or I misread. Wait, the original expression is $(2^{-5} \times 2^{2})^4$. Let's compute step by step:

First, inside the parentheses: $2^{-5} \times 2^{2} = 2^{-5 + 2} = 2^{-3}$. Then raise to the 4th power: $(2^{-3})^4 = 2^{-12}$. To write with a positive exponent, we use the rule $a^{-n} = \frac{1}{a^n}$, so $2^{-12} = \frac{1}{2^{12}}$. But $2^{12} = 4096$, so $\frac{1}{4096}$, but the problem says "exponential form", so probably $\frac{1}{2^{12}}$ or $2^{-12}$, but the problem specifies "positive exponent". So $2^{-12} = \frac{1}{2^{12}}$, which is exponential form with a positive exponent in the denominator. So the answer is $\frac{1}{2^{12}}$ or $2^{-12}$, but since it needs positive exponent, $\frac{1}{2^{12}}$. Wait, but maybe the problem allows the base with a positive exponent, so $2^{-12} = \frac{1}{2^{12}}$, so the answer is $\frac{1}{2^{12}}$ or $2^{-12}$, but let's check the exponent rules again.

Wait, maybe I made a mistake in step 1. Let's re-express:

$(2^{-5} \times 2^{2})^4$

First, multiply the exponents inside the parentheses: $2^{-5 + 2} = 2^{-3}$ (product rule: $a^m \times a^n = a^{m + n}$)

Then, raise to the 4th power: $(2^{-3})^4 = 2^{-3 \times 4} = 2^{-12}$ (power of a power rule: $(a^m)^n = a^{m \times n}$)

Now, to convert to positive exponent: $2^{-12} = \frac{1}{2^{12}}$ (negative exponent rule: $a^{-n} = \frac{1}{a^n}$)

So the simplified form with a positive exponent is $\frac{1}{2^{12}}$ or $2^{-12}$ (but $2^{-12}$ has a negative exponent, so we need to write it as $\frac{1}{2^{12}}$).

Answer:

$\frac{1}{2^{12}}$ (or $2^{-12}$ but since it needs positive exponent, $\frac{1}{2^{12}}$)