Question

Hello, is it possible to show me the steps to solve this answer? I am unsure as to why I got it wrong.

Answer 1

Given:

$(\frac{5e^{-3}}{2e^{18}e^{-13}})^{-4}$

Required:

We need to simplify the given expression.

Explanation:

$Use\text{ }(\frac{a}{b})^n=\frac{a^n}{b^n}.$

$(\frac{5e^{-3}}{2e^{18}e^{-13}})^{-4}=\frac{(5e^{-3})^{-4}}{(2e^{18}e^{-13})^{-4}}$$Use\text{ }(ab)^n=a^nb^n,\text{ and }(a^n)^m=a^{n\times m}.$

$(\frac{5e^{-3}}{2e^{18}e^{-13}})^{-4}=\frac{5^{-4}e^{-3\times(-4)}}{{2^{-4}e^{18\times(-4)}e^{-13\times(-4)}}}$$(\frac{5e^{-3}}{2e^{18}e^{-13}})^{-4}=\frac{5^{-4}e^{12}}{{2^{-4}e^{-72}e^{52}}}$

$Use\text{ }a^na^m=a^{n+m}.$

$(\frac{5e^{-3}}{2e^{18}e^{-13}})^{-4}=\frac{5^{-4}e^{12}}{{2^{-4}e^{-72+52}}}$

$(\frac{5e^{-3}}{2e^{18}e^{-13}})^{-4}=\frac{5^{-4}e^{12}}{{2^{-4}e^{-20}}}$$Use\text{ }\frac{1}{a^{-n}}=a^n\text{ and }a^{-n}=\frac{1}{a^n}.$

$(\frac{5e^{-3}}{2e^{18}e^{-13}})^{-4}=\frac{2^4e^{12}e^{20}}{{5^4}}$$(\frac{5e^{-3}}{2e^{18}e^{-13}})^{-4}=\frac{2^4e^{12+20}}{{5^4}}$$(\frac{5e^{-3}}{2e^{18}e^{-13}})^{-4}=\frac{2^4e^{32}}{{5^4}}$$(\frac{5e^{-3}}{2e^{18}e^{-13}})^{-4}=\frac{16e^{32}}{625}$

Final answer:

$(\frac{5e^{-3}}{2e^{18}e^{-13}})^{-4}=\frac{16e^{32}}{625}$

Alternate method:

$(\frac{5e^{-3}}{2e^{18}e^{-13}})^{-4}=(\frac{2e^{18}e^{-13}}{5e^{-3}})^4$

$(\frac{5e^{-3}}{2e^{18}e^{-13}})^{-4}=(\frac{2e^{18-13}}{5e^{-3}})^4$

$(\frac{5e^{-3}}{2e^{18}e^{-13}})^{-4}=(\frac{2e^5}{5e^{-3}})^4$

$(\frac{5e^{-3}}{2e^{18}e^{-13}})^{-4}=(\frac{2e^5e^3}{5})^4$

$(\frac{5e^{-3}}{2e^{18}e^{-13}})^{-4}=(\frac{2e^{5+3}}{5})^4$

$(\frac{5e^{-3}}{2e^{18}e^{-13}})^{-4}=(\frac{2e^8}{5})^4$

Final answer:

$(\frac{5e^{-3}}{2e^{18}e^{-13}})^{-4}=\frac{16e^{32}}{625}$