Question

10 which of the following two values are equal to 1? a \\(dfrac{(8^2)^3}{8^5}\\) b \\(dfrac{(8^2)^3}{8^6}\\) c \\(8^{-3} \times 8^4 \times 8\\) d \\(8^3 \times 8^{-4} \times 8\\) e \\((8^2)^{-2}\\)

Explanation:

Step1: Analyze Option A

Use exponent rule \((a^m)^n = a^{mn}\) and \(\frac{a^m}{a^n}=a^{m - n}\).
\(\frac{(8^{2})^{3}}{8^{5}}=\frac{8^{6}}{8^{5}} = 8^{6 - 5}=8^{1}=8
eq1\)

Step2: Analyze Option B

Use exponent rule \((a^m)^n = a^{mn}\) and \(\frac{a^m}{a^n}=a^{m - n}\).
\(\frac{(8^{2})^{3}}{8^{6}}=\frac{8^{6}}{8^{6}} = 8^{6 - 6}=8^{0}=1\)

Step3: Analyze Option C

Use exponent rule \(a^m\times a^n=a^{m + n}\).
\(8^{-3}\times8^{4}\times8=8^{-3 + 4+1}=8^{2}=64
eq1\)

Step4: Analyze Option D

Use exponent rule \(a^m\times a^n=a^{m + n}\).
\(8^{3}\times8^{-4}\times8=8^{3-4 + 1}=8^{0}=1\)

Step5: Analyze Option E

Use exponent rule \((a^m)^n = a^{mn}\).
\((8^{2})^{-2}=8^{-4}=\frac{1}{8^{4}}
eq1\)

Answer:

B. \(\boldsymbol{\frac{(8^{2})^{3}}{8^{6}}}\), D. \(\boldsymbol{8^{3}\times8^{-4}\times8}\)