Question

Find x such that ${\left(\frac{3}{5}\right)}^3\times {\left(\frac{3}{5}\right)}^{-6}={\left(\frac{3}{5}\right)}^{2x-1}$.

Answer 1

${\left(\frac{3}{5}\right)}^3\times {\left(\frac{3}{5}\right)}^{-6}={\left(\frac{3}{5}\right)}^{2x-1}$
⇒ ${\left(\frac{3}{5}\right)}^{3+(-6)}={\left(\frac{3}{5}\right)}^{2x-1}$ $\left[s\mathrm{ince}{\mathrm{a}}^{\mathrm{m}}\times {\mathrm{a}}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{m}+\mathrm{n}}\right]$
⇒ ${\left(\frac{3}{5}\right)}^{-3}={\left(\frac{3}{5}\right)}^{2x-1}$
On equating the exponents:
−3 = 2x − 1
⇒ 2x = −3 + 1
⇒ 2x = −2
∴ x = $\left(\frac{-2}{2}\right)=-1$