How do you solve $ln(x)=3+ln(x-3)$ ?

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Answer 1 · 2015-08-13T00:21:45

$color(red)( x=(3e^3)/(e^3-1))$

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

Subtract $ln(x-3)$ from each side.

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

Recall that $lna-lnb=ln(a/b)$ .

∴ $ln(x/(x-3)) = 3$

Convert the logarithmic equation to an exponential equation.

$e^ln(x/(x-3)) = e^3$

Remember that $e^lnx =x$ , so

$x/(x-3)=e^3$

$x=e^3(x-3) = xe^3-3e^3$

$xe^3-x=3e^3$

$x(e^3-1)=3e^3$

$x=(3e^3)/(e^3-1)$

Check:

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

If $x=(3e^3)/(e^3-1)$ ,

$ln((3e^3)/(e^3-1)) = 3+ln((3e^3)/(e^3-1)-3)$

$ln(3e^3)-ln(e^3-1)=3+ln((3e^3-3(e^3-1))/(e^3-1))$

$ln(3e^3)-ln(e^3-1)=3+ln((color(red)(cancel(color(black)(3e^3)))-color(red)(cancel(color(black)(3e^3)))+3)/(e^3-1))$

$ln(3e^3)-color(red)(cancel(color(black)(ln(e^3-1))))=3+ln3-color(red)(cancel(color(black)(ln(e^3-1))))$

$ln3+3=3+ln3$

$x=(3e^3)/(e^3-1)$ is a solution.

How do you solve ln(x)=3+ln(x-3)ln(x)=3+ln(x-3)?