Solve the following equations for $x$:$ 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 find the value of $x$.

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,

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

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

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

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

$(2)^{2x-2-3+2x}=(2)^{-3x}$

$(2)^{4x-5}=(2)^{-3x}$

Comparing the powers on both sides, we get,

$4x-5=-3x$

$4x+3x=5$

$7x=5$

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

Therefore, the value of $x$ is $\frac{5}{7}$.  

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Updated on: 10-Oct-2022

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