Question

Find the value of $x$
${[{\left(\frac{2}{5}\right)}^{-6}÷{\left(\frac{2}{5}\right)}^3]}^x={\left(\frac{2}{5}\right)}^{-9}$

Answer 1

${[{\left(\frac{2}{5}\right)}^{-6}÷{\left(\frac{2}{5}\right)}^3]}^x={\left(\frac{2}{5}\right)}^{-9}$

${[{\left(\frac{2}{5}\right)}^{-6}\times {\left(\frac{5}{2}\right)}^3]}^x={\left(\frac{2}{5}\right)}^{-9}$ since $a^m=\frac{1}{a^{-m}}$

${[{\left(\frac{5}{2}\right)}^6\times {\left(\frac{5}{2}\right)}^3]}^x={\left(\frac{2}{5}\right)}^{-9}$ since $a^m=\frac{1}{a^{-m}}$

${\left[{\left(\frac{5}{2}\right)}^{6+3}\right]}^x={\left(\frac{5}{2}\right)}^9$ since $a^m\times a^n=a^{m+n}$

${\left[{\left(\frac{5}{2}\right)}^9\right]}^x={\left(\frac{5}{2}\right)}^9$ since $a^m=\frac{1}{a^{-m}}$

${\left(\frac{5}{2}\right)}^{9x}={\left(\frac{5}{2}\right)}^9$ since ${\left(a^m\right)}^n=a^{mn}$

Since bases are the same we can equate the powers

$\Rightarrow 9x=9$

$\Rightarrow x=\frac{9}{9}=1$

$\therefore x=1$