Question

Find the value of $x$ if ${0.6}^x{\left(\frac{25}{9}\right)}^{x^2-12}={\left(\frac{27}{125}\right)}^3$

Answer 1

$\Rightarrow {\left(\frac{6}{10}\right)}^x{\left(\frac{25}{9}\right)}^{x^2-12}={\left(\frac{27}{125}\right)}^3$

$\Rightarrow {\left(\frac{3}{5}\right)}^x{\left(\frac{5^2}{3^2}\right)}^{x^2-12}={\left(\frac{3^3}{5^3}\right)}^3$

$\Rightarrow {\left(\frac{3}{5}\right)}^x{\left(\frac{5}{3}\right)}^{2x^2-24}={\left(\frac{3}{5}\right)}^9$

$\Rightarrow \frac{3^x}{5^x}\times \frac{5^{2x^2-24}}{3^{2x^2-24}}=\frac{3^9}{5^9}$

$\Rightarrow \frac{3^{x-2x^2+24}}{5^{x-2x^2+24}}=\frac{3^9}{5^9}$

Equating the powers with the same base, we get

$3^{x-2x^2+24}=3^9$ and $5^{x-2x^2+24}=5^9$

$\Rightarrow x-2x^2+24=9$ and $x-2x^2+24=9$

$\Rightarrow -2x^2+x+24-9=0$ and $-2x^2+x+24-9=0$

$\Rightarrow -2x^2+x+15=0$ and $-2x^2+x+15=0$

$\Rightarrow 2x^2-x-15=0$

$\Rightarrow 2x^2-6x+5x-15=0$

$\Rightarrow 2x(x-3)+5(x-3)=0$

$\Rightarrow (x-3)(2x+5)=0$

$\Rightarrow x=3,\frac{-5}{2}$