Question

Solve the following equation for $x$ :

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

Answer 1

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

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

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

$[\because (a^m{)}^n=a^{mn}]$

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

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

$[\because \frac{a^m}{a^n}=\frac{1}{a^{n-m}}]$

$\Rightarrow \frac{1}{2^{5-4x}}=\frac{1}{2^{3x}}$

On comparing both sides, we get

$\Rightarrow 5-4x=3x$

$\Rightarrow 4x+3x=5$

$\Rightarrow 7x=5$

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