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

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

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Answer 1 · 2015-07-08T04:09:52

I found $x = 7$ $x=7$

Explanation:

You can use one rule of the logs:
$log a - log b = \frac{log a}{b}$ $loga-logb=loga/b$
to get:
$ln (\frac{x + 5}{3}) = ln (x - 3)$ $ln((x+5)/3)=ln(x-3)$
take the exponential of both sides:
$e^{ln (\frac{x + 5}{3})} = e^{ln (x - 3)}$ $e^(ln((x+5)/3))=e^(ln(x-3))$
that gives you (cancelling $ln$ $ln$ and $e$ $e$ ):
$\frac{x + 5}{3} = x - 3$ $(x+5)/3=x-3$
$x + 5 = 3 x - 9$ $x+5=3x-9$
$2 x = 14$ $2x=14$
$x = 7$ $x=7$

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

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Explanation:

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

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