If $(-7)^x \times\left(\frac{1}{-7}\right)^{-3}=7^8$, then value of $x$ is $…………..$
Answer
If $(-7)^x \times\left(\frac{1}{-7}\right)^{-3}=7^8$, then value of $x$ is$......$
If $(-7)^x \times\left(\frac{1}{-7}\right)^{-3}=7^8$
$\Rightarrow (-7)^x \times\left(\frac{1}{-7}\right)^{-3}=(-7)^8 \quad\{8$ is even$\}$
$\Rightarrow (-7)^x \times(-7)^{+3}=(-7)^8$
$\Rightarrow (-7)^{x+3}=(-7)^8$
Comparing, we get
$x+3=8$
$\Rightarrow x=8-3=5$
$\therefore x=5$