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jmc

algebra intermediate

Problem

Find the solution to the equation

lo g_{3 x} 4 = lo g_{2 x} 8

, where

x

is a positive real number other than

\frac{1}{3}

or

\frac{1}{2}

.

Solution

By the change-of-base formula,

\frac{lo g 4}{lo g 3 x} = \frac{lo g 8}{lo g 2 x} .

Then

\frac{lo g 3 x}{lo g 2 ^{2}} = \frac{lo g 2 x}{lo g 2 ^{3}},

so

\frac{lo g x + lo g 3}{2 lo g 2} = \frac{lo g x + lo g 2}{3 lo g 2} .

Hence,

\frac{lo g x + lo g 3}{2} = \frac{lo g x + lo g 2}{3},

so

3 lo g x + 3 lo g 3 = 2 lo g x + 2 lo g 2.

Then

lo g x = 2 lo g 2 - 3 lo g 3 = lo g 4 - lo g 27 = lo g \frac{4}{27},

so

x = \frac{4}{27} .

Final answer

\frac{4}{27}