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

Given:

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

To do: 

We have to solve the given equation.

Solution:

We know that,

$(a^{m})^{n}=a^{m n}$

$a^{m} \times a^{n}=a^{m+n}$

$a^{m} \div a^{n}=a^{m-n}$

$a^{0}=1$

Therefore,

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

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

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

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

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

Comparing both sides, we get,

$4 x-5=-3 x$

$\Rightarrow 4 x+3 x=5$

$\Rightarrow 7 x=5$

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

The values of $x$ is $\frac{5}{7}$.

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Updated on: 10-Oct-2022

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