# Solve For X: Log3 X - Log32 = 1

Given:

$log_{3}\; x – log_{3} \;2\; = \;1$

We need to solve for x.

$log_{3}\; x\; = \;1\; + \;log_{3}\; 2.$

Since, 1 can be written as $log_{3}\;3\; = \;1$ $\Rightarrow log_{3} \;(3 * 2)\\\Rightarrow log_{3} \;6\\\Rightarrow xv = \;6$

Therefore, x = 6.

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