Question

Question 99(i)

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

Answer 1

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

Using the law of exponents,
$a^m\times a^n=(a{)}^{m+n}$

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

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


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

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

Since, the base is same on both sides, we can equate the exponents.

$\Rightarrow -16=8x$
$\Rightarrow x=-2$