4^{x-1} = (√2)^x | Indices - Solve | SciPiPupil

Question: Solve: $4^{x-1} = ( \sqrt{2}^x)$

Solution:

Given,

$4^{x-1} = (\sqrt{2}^x)$

$or, \left ( 2^2 \right )^{x-1} = 2^{\dfrac{x}{2}}$

$or,  2^{2(x-1)} = 2^{\dfrac{x}{2}}$

$or, 2(x-1) = \dfrac{x}{2}$

$or, 2 × (2x -2) = \dfrac{2 × x}{2}$

$or, 4x -4 = x$

$or, 4x - x -4 +4 = x -x +4$

$or, 3x = 4$

$\therefore, x = \dfrac{4}{3}$

= Answer

Related Notes and Solutions:

Here is the website link to the notes of Indices.

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