Question

Question 99(ii)

$(-2{)}^3\times (-2{)}^6=(-2{)}^{2x-1}$

Answer 1

We have $(-2{)}^3\times (-2{)}^6=(-2{)}^{2x-1}$

Using law of exponents,
$a^{-m}=\frac{1}{a^m}$

Then,$(-2{)}^3(-2{)}^{-6}=(-2{)}^{2x-1}$

$\Rightarrow (-2{)}^{3+(-6)}=(-2{)}^{2x-1}$

$\Rightarrow (-2{)}^{3-6}=(-2{)}^{2x-1}$

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

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

$-3=2x-1\Rightarrow 2x=-3+1$
$\Rightarrow 2x=-2$
$\Rightarrow x=-1$