Question

${\frac{-2}{5}}^{8} \times {\frac{-2}{5}}^{4}={\frac{-2}{5}}^{x-2}$
Find the Value of X​

Answer 1

$\rm {\dfrac{-2}{5}}^8 \times {\dfrac{-2}{5}}^4={\dfrac{-2}{5}}^{x-2}$

$\rm\implies {\dfrac{-2}{5}}^{8+4}={\dfrac{-2}{5}}^{x-2}\qquad ...[since , \ a^m \times a^n = a^{m+n}]$

Bases are same, so bases will be eliminated.

$\rm\implies 8+4=x-2$

$\rm\implies 12=x-2$

$\rm \implies 12+2=x$

$\rm \implies 14=x$

$\qquad\qquad\bigstar\boxed{\bf x=14}\bigstar$

Hence, the value of x is 14

Answer 2

Answer:

the bases are same so add exponents

8+4=x-2

12=x-2

12+2=x (here the minus value goes to another side becomes plus

14=x

x=4

hope it helps you....

thankyou..