How do you solve $2^{x + 3} = 3^{x - 4}$ $2^(x+3) = 3^(x-4)$ ?

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Answer 1 · 2016-09-24T18:59:05

$x = 16.4695$ $x = 16.4695$

As the bases are different, we cannot just compare them.

The variables are in the exponents, so logs are called for.

Log both sides:

${log 2}^{x + 3} = {log 3}^{x - 4} \leftarrow$ $log2^(x+3) = log3^(x-4)" "larr$ use the log power law

$(x + 3) log 2 = (x - 4) log 3 \leftarrow$ $(x+3)log2 = (x-4)log3 " "larr$ move the log terms to one side

$\frac{x + 3}{x - 4} = \frac{log 3}{log 2} = 1.58496$ $(x+3)/(x-4) = (log3)/(log2) = 1.58496$

$\frac{x + 3}{x - 4} = 1.58496 \leftarrow$ $(x+3)/(x-4) = 1.58496" "larr$ cross-mulitply

$x + 3 = 1.58496 (x - 4)$ $x+3 = 1.58496(x-4)$

$x + 3 = 1.58496 x - 6.33985 \leftarrow$ $x +3 = 1.58496x -6.33985" "larr$ re-arrange the terms

$3 + 6.33985 = 1.58496 x - x$ $3+6.33985 = 1.58496x -x$

$9.633985 = 0.58496 x$ $9.633985 = 0.58496x$

$\frac{9.633985}{0.58496} = x$ $9.633985/0.58496 = x$

$x = 16.4695$ $x = 16.4695$

How do you solve 2^(x+3) = 3^(x-4)2x+3=3x−42^(x+3) = 3^(x-4)?