Question #082ad

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Answer 1 · 2017-08-31T04:06:12

$x = 9 e^{2} + 3$ $x=9e^2+3$

Condense the logarithms using the rules:

Then

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

$ln (x - 3) - ln (3^{2}) = 2$ $ln(x-3)-ln(3^2)=2$

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

Rewrite by exponentiating both sides with base $e$ $e$ (that is, undo the logarithm):

$\frac{x - 3}{9} = e^{2}$ $(x-3)/9=e^2$

From here, solving for $x$ $x$ is simple:

$x = 9 e^{2} + 3$ $x=9e^2+3$