Question

evaluate the expression.
\frac{(3^{-2} \cdot 8^{0})^{4}}{3^{-5}}
use the number pad to enter your answer in the box.

Explanation:

Step1: Simplify \(8^0\)

Any non - zero number to the power of 0 is 1, so \(8^0 = 1\). The expression becomes \(\frac{(3^{-2}\times1)^{4}}{3^{-5}}=\frac{(3^{-2})^{4}}{3^{-5}}\)

Step2: Apply the power of a power rule \((a^{m})^{n}=a^{mn}\)

For \((3^{-2})^{4}\), using the rule \((a^{m})^{n}=a^{mn}\), we have \(3^{-2\times4}=3^{-8}\). Now the expression is \(\frac{3^{-8}}{3^{-5}}\)

Step3: Apply the quotient rule for exponents \(\frac{a^{m}}{a^{n}}=a^{m - n}\)

Using the rule \(\frac{a^{m}}{a^{n}}=a^{m - n}\) with \(a = 3\), \(m=-8\) and \(n = - 5\), we get \(3^{-8-(-5)}=3^{-8 + 5}=3^{-3}\)

Step4: Rewrite \(3^{-3}\) using the negative exponent rule \(a^{-n}=\frac{1}{a^{n}}\)

\(3^{-3}=\frac{1}{3^{3}}=\frac{1}{27}\) Wait, no, wait. Wait, let's re - check step 3. Wait, \(\frac{3^{-8}}{3^{-5}}=3^{-8-(-5)}=3^{-8 + 5}=3^{-3}\)? Wait, no, that's wrong. Wait, \(-8-(-5)=-8 + 5=-3\)? Wait, no, wait the original numerator after step 2 is \((3^{-2}\times8^{0})^{4}\). Since \(8^{0}=1\), \((3^{-2})^{4}=3^{-8}\). The denominator is \(3^{-5}\). So \(\frac{3^{-8}}{3^{-5}}=3^{-8-(-5)}=3^{-3}\)? Wait, no, that can't be. Wait, no, let's start over.

Wait, step 1: \(8^0 = 1\), so the numerator inside the big fraction is \((3^{-2}\times1)^{4}=(3^{-2})^{4}\). Then \((3^{-2})^{4}=3^{-8}\) (because \((a^{m})^{n}=a^{mn}\), so \(-2\times4=-8\)). Then the fraction is \(\frac{3^{-8}}{3^{-5}}\). Now, the rule for dividing exponents with the same base is \(a^{m}\div a^{n}=a^{m - n}\). So \(3^{-8}\div3^{-5}=3^{-8-(-5)}=3^{-8 + 5}=3^{-3}\)? Wait, no, \(-8-(-5)=-8 + 5=-3\). But \(3^{-3}=\frac{1}{3^{3}}=\frac{1}{27}\)? Wait, that seems wrong. Wait, maybe I made a mistake in the sign. Wait, let's re - examine the original expression:

Original expression: \(\frac{(3^{-2}\cdot8^{0})^{4}}{3^{-5}}\)

First, \(8^{0}=1\), so inside the numerator: \((3^{-2}\times1)^{4}=(3^{-2})^{4}\)

\((3^{-2})^{4}=3^{-8}\) (power of a power: multiply exponents)

Now, the fraction is \(\frac{3^{-8}}{3^{-5}}\)

Using the rule \(\frac{a^{m}}{a^{n}}=a^{m - n}\), so \(m=-8\), \(n = - 5\)

\(m - n=-8-(-5)=-8 + 5=-3\), so \(3^{-3}=\frac{1}{3^{3}}=\frac{1}{27}\)? Wait, but that seems incorrect. Wait, no, wait, maybe I messed up the order of operations. Wait, let's do it again.

Wait, the original expression is \(\frac{(3^{-2}\times8^{0})^{4}}{3^{-5}}\)

First, simplify the numerator:

\(8^{0}=1\), so \(3^{-2}\times8^{0}=3^{-2}\times1 = 3^{-2}\)

Then raise to the 4th power: \((3^{-2})^{4}=3^{-8}\) (because \((a^{m})^{n}=a^{mn}\), so \(-2\times4=-8\))

Now, the denominator is \(3^{-5}\)

So we have \(\frac{3^{-8}}{3^{-5}}\)

Using the exponent rule \(\frac{a^{m}}{a^{n}}=a^{m - n}\), so \(3^{-8-(-5)}=3^{-3}\)? Wait, no, \(-8-(-5)=-8 + 5=-3\), so \(3^{-3}=\frac{1}{27}\). But let's check with positive exponents.

Alternatively, rewrite \(3^{-8}=\frac{1}{3^{8}}\) and \(3^{-5}=\frac{1}{3^{5}}\)

Then \(\frac{\frac{1}{3^{8}}}{\frac{1}{3^{5}}}=\frac{1}{3^{8}}\times3^{5}=\frac{3^{5}}{3^{8}}=3^{5 - 8}=3^{-3}=\frac{1}{27}\). Wait, but that's the same result. Wait, but maybe I made a mistake in the sign when applying the power of a power. Wait, no, \((3^{-2})^{4}=3^{-8}\) is correct because \((a^{m})^{n}=a^{mn}\), \(-2\times4=-8\). Then dividing by \(3^{-5}\) is the same as multiplying by \(3^{5}\), so \(3^{-8}\times3^{5}=3^{-8 + 5}=3^{-3}=\frac{1}{27}\). Yes, that's correct.

Wait, but let's check with another approach. Let's use positive exponents from the start.

\(3^{-2}=\frac{1}{3^{2}}=\frac{1}{9}\), \(8^{0}=1\), so \((3^{-2}\times8^{0})^{4}=(\frac{1}{9}\times1)^{4}=(\frac{1}{9})^{4}=\frac{1}{9^{4}}=\frac{1}{6561}\)

\(3^{-5}=\frac{1}{3^{5}}=\frac{1}{243}\)

Then \(\frac{\frac{1}{6561}}{\frac{1}{243}}=\frac{1}{6561}\times243=\frac{243}{6561}=\frac{1}{27}\) (because \(6561\div243 = 27\))

Yes, so the result is \(\frac{1}{27}\)

Answer:

\(\frac{1}{27}\)