How do you solve $2^{3 x + 1} = 3^{x - 2}$ $2^(3x+1)=3^(x-2)$ ?

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Answer 1 · 2016-07-09T05:26:02

$- \frac{log 18}{log (\frac{3}{8})} = - 2.947$ $- log 18/log (3/8)=-2.947$ , nearly,

The given equation can be simplified as

$\frac{{(2^{3})}^{x}}{3^{x}} = \frac{3^{- 2}}{2^{1}}$ $(2^3)^x/3^x=3^(-2)/2^1$ .

So, ${(\frac{3}{8})}^{x} = 18$ $(3/8)^x=18$ .

Equating logarithms and solving,

$x = \frac{log 18}{log (\frac{3}{8})}$ $x=log 18/log (3/8)$

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