Question

Solve the following equations:

v) $4^{x-1}\times (0.5{)}^{3-2x}={\left(\frac{1}{8}\right)}^x$

Answer 1

Given : $4^{x-1}\times (0.5{)}^{3-2x}={\left(\frac{1}{8}\right)}^x$

$\Rightarrow (2^2{)}^{x-1}\times {\left(\frac{1}{2}\right)}^{3-2x}={\left(\frac{1}{2^3}\right)}^x$

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

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

$\Rightarrow 2^{\left[\right(2x-2)-(3-2x\left)\right]}=2^{-3x}$

$\Rightarrow 2^{[2x-2-3+2x]}=2^{-3x}$

$\Rightarrow 2^{4x-5}=2^{-3x}$

On comparing both the sides, we get

$\Rightarrow 4x-5=-3x$

$\Rightarrow 4x+3x=5$

$\Rightarrow 7x=5$

$\therefore x=\frac{5}{7}$