How do you solve $Ln (y) – Ln (y-4) = Ln (2)$ ?

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Answer 1 · 2016-03-10T22:03:27

$y=8$

putting everything on one side we get

$lny-ln(y-4)-ln2=0$

Use the property that $log_b (x/y)=log_b x-log_by$

$ln(y/(2(y-4)))=0$

Take the inverse log, in this case the inverse of $ln(x)$ is $e^x$

$e^0=y/(2(y-4))$

$1=y/(2(y-4))$

$2y-8=y$

$2y-y=8$

$y=8$

How do you solve Ln (y) – Ln (y-4) = Ln (2) Ln (y) – Ln (y-4) = Ln (2) ?