Exponent Law: Negative Exponents Division

Question

I'm trying to relearn high-school maths after years of decay. This is a very basic exponent law question.

How can I prove the following in a step-by-step fashion:

$$ \biggl(\frac{7^3}{3^8}\biggl)^{-2} = \frac{3^{16}}{7^6} $$

Thanks!

Answer 1

You can go by as follows:

Using exponentiation rules

$$ \Big(\frac{7^3}{3^8}\Big)^{-2}=\Big(\frac{7^3}{3^8}\Big)^{2\cdot (-1)}= $$ $$ =\Big(\frac{7^{3\cdot 2}}{3^{8\cdot 2}}\Big)^{-1}=\frac{1}{\frac{7^{3\cdot 2}}{3^{8\cdot 2}}}= $$ $$ =\frac{3^{8\cdot 2}}{7^{3\cdot 2}}=\frac{3^{16}}{7^{6}}. $$

Hope this helps.