Question

The value of $x$ satisfying the equation $\left(\frac{2^{x-1}{.4}^{x+1}}{{16}^{x-1}}\right)=64$ is

Answer 1

Given, $\left(\frac{2^{x-1}{.4}^{x+1}}{{16}^{x-1}}\right)=64$

$\left(\frac{2^{x-1}{.2}^{2(x+1)}}{2^4{}^{(x-1)}}\right)=64$

$\left(\frac{2^{x-1}{.2}^{2x+2}}{2^{4x-4}}\right)=64$

$\left(\frac{2^{x-1+2x+2}}{2^{4x-4}}\right)=64$ ($as{a}^m\times a^n=a^{m+n}$)

$\left(\frac{2^{3x+1}}{2^{4x-4}}\right)=64$

$2^{3x+1-(4x-4)}=64$ ($as{a}^m÷a^n=a^{m-n}$)

$2^{3x+1-4x+4}=64$

$2^{-x+5}=64$

we can write $64=2\times 2\times 2\times 2\times 2\times 2=2^6$

So, $2^{-x+5}$ = $2^6$

$-x+5=6$

$-x=6-5$

$-x=1$

Hence, $x=-1$