Question: Solve: $2^{x-3} × 3^{x-4} = 3^{-1}$

Solution:
Given,

$2^{x-3} × 3^{x-4} = 3^{-1}$

$or, 2^x × 2^{-3} × 3^x × 3^{-4} = 3^{-1}$

$or, (2^3)^x × \dfrac{1}{2^3} × \dfrac{1}{3^4} = \dfrac{1}{3}$

$or, 6^x × \dfrac{1 × 3}{8×9×9} = \dfrac{1 × 3}{3}$

$or, 6^x × \dfrac{1}{8×27} = 1$

$or, 6^x = 216$

$or, 6^x = 6^3$

$\therefore x = 3$
= Answer

Related Notes and Solutions:

Here is the website link to the notes of Indices.

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