Question

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

• A. -3
• B. -2
• C. -1
• D. 0

Answer 1

The correct option is B

-2

$Given,{\left(\frac{5}{3}\right)}^{-2}\times {\left(\frac{5}{3}\right)}^{-14}={\left(\frac{5}{3}\right)}^{8x}$
Using law of exponents, $a^m\times a^n=(a{)}^{m+n}$
$\Rightarrow {\left(\frac{5}{3}\right)}^{-2-14}={\left(\frac{5}{3}\right)}^{8x}$
$\Rightarrow {\left(\frac{5}{3}\right)}^{-16}={\left(\frac{5}{3}\right)}^{8x}$
Since the bases are equal, the exponents should also be equal
$\Rightarrow -16=8x$
$\Rightarrow x=-2$