Question

Need help solving -3^4x-3^4/(-3^2)^3

Answer 1

Answer:

The solved given expression is

$\frac{-3^4x-3^4}{(-3^2)^3}=-\frac{(x+1)}{9}$

Step-by-step explanation:

Given expression is $\frac{-3^4x-3^4}{(-3^2)^3}$

To solving the given expression as below :

$\frac{-3^4x-3^4}{(-3^2)^3}$

$\frac{-3^4x-3^4}{(-3^2)3}=\frac{3^4(-x-1)}{(-3^2)3}$

$=\frac{-3^4(x+1)}{(3^2)3}$  ( by using the property $(a^m)^n=a^{mn}$ )

$=\frac{-3^4(x+1)}{3^6}$

$=\frac{-(x+1)}{3^6.3^{-4}}$  ( by using the property $a^m=\frac{1}{a^{-m}}$ )

$=\frac{-(x+1)}{3^{6-4}}$ ( by using the property $a^m.a^n=a^{m+n}$  )

$=-\frac{(x+1)}{3^2}$

$=-\frac{(x+1)}{9}$

Therefore $\frac{-3^4x-3^4}{(-3^2)^3}=-\frac{(x+1)}{9}$

Therefore the solved given expression is

$\frac{-3^4x-3^4}{(-3^2)^3}=-\frac{(x+1)}{9}$