The value of x in the given equation {4^x} - | vedclass

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Question

(b) Equation, ${4^x} - {3^{x - \frac{1}{2}}} = {3^{x + \frac{1}{2}}} - {2^{2x - 1}}$

==> ${2^{2x}} + {2^{2x - 1}} = {3^{x + \frac{1}{2}}} + {3^{

Vedclass · Accepted Answer

$\frac{3}{2}$

Vedclass · Answer

(b) Equation, ${4^x} - {3^{x - \frac{1}{2}}} = {3^{x + \frac{1}{2}}} - {2^{2x - 1}}$

==> ${2^{2x}} + {2^{2x - 1}} = {3^{x + \frac{1}{2}}} + {3^{x - \frac{1}{2}}}$

==> ${2^{2x}}\left( {1 + \frac{1}{2}} ight) = {3^{x - \frac{1}{2}}}(1 + 3)$

==> ${2^{2x}}.\frac{3}{2} = {3^{x - \frac{1}{2}}}.4$ 

 ==> ${2^{2x - 3}} = {3^{x - \frac{3}{2}}}$

Taking log both sides

==> $(2x - 3)\log 2 = (x - 3/2)\log 3$

==> $2x\log 2 - 3\log 2 = x\log 3 - \frac{3}{2}\log 3$

==> $x\log 4 - x\log 3 = 3\log 2 - \frac{3}{2}\log 3$

==> $x\log \left( {\frac{4}{3}} ight) = \log 8 - \log 3\sqrt 3 $

==>${\left( {\frac{4}{3}} ight)^x} = \frac{8}{{3\sqrt 3 }}$

$&rArr; {\left( {\frac{4}{3}} ight)^x} = {\left( {\frac{4}{3}} ight)^{3/2}}$

$	herefore \,\,x = \frac{3}{2}$

Trick: Cheak the equation with options then only option $(b)$ satisfies the equation.

The value of $x$ in the given equation ${4^x} - {3^{x\,\; - \;\frac{1}{2}}} = {3^{x + \frac{1}{2}}} - {2^{2x - 1}}$is

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