How do you solve $ln (x - 3) - ln (x - 5) = ln 5$ $ln(x-3)-ln(x-5)=ln5$ ?

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Answer 1 · 2016-07-22T06:24:01

$x = \frac{11}{2}$ $x=11/2$

The solution must be:

$x - 3 > 0 and x - 5 > 0$ $x-3>0 and x-5>0$

that's $x > 5$ $x>5$

Since $log a - log b = log (\frac{a}{b})$ $loga-logb=log(a/b)$

the given equation is equivalent to:

$ln (\frac{x - 3}{x - 5}) = ln 5$ $ln((x-3)/(x-5))=ln5$

$\frac{x - 3}{x - 5} = 5$ $(x-3)/(x-5)=5$

$x - 3 = 5 (x - 5)$ $x-3=5(x-5)$

$x - 3 = 5 x - 25$ $x-3=5x-25$

$4 x = 22$ $4x=22$

$x = \frac{11}{2}$ $x=11/2$

that's >5

How do you solve ln(x-3)-ln(x-5)=ln5ln(x−3)−ln(x−5)=ln5ln(x-3)-ln(x-5)=ln5?