Question

Find $n$, so that
$a)$ ${\left(\frac{4}{5}\right)}^3\times {\left(\frac{4}{5}\right)}^{-6}={\left(\frac{4}{5}\right)}^{2n-1}$
$b)$ $n\left(3^{-5}\right)=3$
$c)$ $n{\left(-5\right)}^4÷n^2=5$

Answer 1

$a)$ ${\left(\frac{4}{5}\right)}^3\times {\left(\frac{4}{5}\right)}^{-6}={\left(\frac{4}{5}\right)}^{2n-1}$
$\Rightarrow {\left(\frac{4}{3}\right)}^{3-6}={\left(\frac{4}{5}\right)}^{2n-1}\left[\because a^m\times a^n=a^{m+n}\right]$
Comparing powers
$\Rightarrow$$3-6=2n-1$
$\Rightarrow$$3-6+1=2n$
$\Rightarrow$$4-6=2n$
$\Rightarrow$$-2=2n$
$n=-1$
$b)$ $n\left(3^{-5}\right)=3$
$\Rightarrow$$n\times {\left(\frac{1}{3}\right)}^5=3\left[\because a^{-n}=\frac{1}{a^n}\right]$
$\Rightarrow$$n=3\times 3^5$
$n=3^6$
$c)$ $n{\left(-5\right)}^4÷n^2=5$
$\Rightarrow$$n{\left(-5\right)}^4\times \frac{1}{n^2}=5$
$\Rightarrow$${\left(-5\right)}^4\times \frac{1}{n}=5$
$\Rightarrow$${\left(-5\right)}^4\times \frac{1}{5}=n$
$\Rightarrow$$5\times 5\times 5\times 5\times \frac{1}{5}=n$
$n=125$