How can I find X? (Exponential Equation)

Question

$2^{3 x + 1} = 3^{x - 2}$ $2^(3x+1) = 3^(x-2)$

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Answer 1 · 2017-04-13T13:13:20

$x = - \frac{ln 2 + 2 ln 3}{3 ln 2 - ln 3}$ $x = -(ln 2 + 2 ln 3)/(3 ln 2 - ln 3)$

Given:

$2^{3 x + 1} = 3^{x - 2}$ $2^(3x+1) = 3^(x-2)$

Take logs of both sides to get:

$(3 x + 1) ln 2 = (x - 2) ln 3$ $(3x+1)ln 2 = (x-2)ln3$

Multiply out to get:

$(3 ln 2) x + ln 2 = (ln 3) x - 2 ln 3$ $(3 ln 2)x + ln 2 = (ln 3)x- 2 ln 3$

Subtract $ln 2 + (ln 3) x$ $ln 2 + (ln 3)x$ from both sides to get:

$(3 ln 2 - ln 3) x = - ln 2 - 2 ln 3$ $(3 ln 2 - ln 3)x = - ln 2 - 2 ln 3$

Divide both sides by $(3 ln 2 - ln 3)$ $(3 ln 2 - ln 3)$ to get:

$x = - \frac{ln 2 + 2 ln 3}{3 ln 2 - ln 3}$ $x = -(ln 2 + 2 ln 3)/(3 ln 2 - ln 3)$