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

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Answer 1 · 2016-09-24T18:59:05

$x = 16.4695$

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

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

Log both sides:

$log2^(x+3) = log3^(x-4)" "larr$ use the log power law

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

$(x+3)/(x-4) = (log3)/(log2) = 1.58496$

$(x+3)/(x-4) = 1.58496" "larr$ cross-mulitply

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

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

$3+6.33985 = 1.58496x -x$

$9.633985 = 0.58496x$

$9.633985/0.58496 = x$

$x = 16.4695$

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