Question

Find the value of x so that ${\displaystyle {\left(\frac{5}{3}\right)}^{-2}\times {\left(\frac{5}{3}\right)}^{-14}={\left(\frac{5}{3}\right)}^{8x}}$

Answer 1

We have, ${\displaystyle {\left(\frac{5}{3}\right)}^{-2}\times {\left(\frac{5}{3}\right)}^{-14}={\left(\frac{5}{3}\right)}^{8x}}$

Using law of exponents, a^m × aⁿ = (a)^{m + n}  ...[∵ a is non-zero integer]

Then, ${\displaystyle {\left(\frac{5}{3}\right)}^{-2}\times {\left(\frac{5}{3}\right)}^{-14}={\left(\frac{5}{3}\right)}^{8x}}$

⇒ ${\displaystyle {\left(\frac{5}{3}\right)}^{-2-14}={\left(\frac{5}{3}\right)}^{8x}}$

⇒ ${\displaystyle {\left(\frac{5}{3}\right)}^{-16}={\left(\frac{5}{3}\right)}^{8x}}$

On comparing both sides, we get

16 = 8x

⇒ x = – 2