Question: Solve: $2^{3x-5}.a^{x-2} = 2^{x-2}.a^{1-x}$

Solution:
Given,

$or, 2^{3x-5}.a^{x-2} = 2^{x-2}.a^{1-x}$

$or, 2^{3x} × 2^{-5} × a^x × a^{-2} = 2^x ×2^{-2} × a^1 × a^{-x}$

$or, \dfrac{2^{3x}}{2^5} × \dfrac{a^x}{a^2} = \dfrac{2^x}{2^2} × \dfrac{a}{a^{x}}$

$or, \dfrac{2^{3x} × a^x}{2^5 × a^2} = \dfrac{2^x × a}{2^2 × a^x}$

$or, \dfrac{2^{3x} × a^x × a^x}{2^x} = \dfrac{2^5 × a^2 × a}{2^2}$

$or, 2^{3x -x} × a^{x +x} = a^{1+2} × 2^{5-2}$

$or, 2^{2x} × a^{2x} = a^3 × 2^3$

$or, (2a)^{2x} = (2a)^{3}$

$or, 2x = 3$

$\therefore, x = \dfrac{3}{2}$


Related Notes and Solutions:

Here is the website link to the notes of Indices.

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