How do you solve $ln (x) - ln (x - 1) = ln (3)$ $Ln (x) - Ln (x-1) = Ln (3)$ ?

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How do you solve $ln (x) - ln (x - 1) = ln (3)$ $Ln (x) - Ln (x-1) = Ln (3)$ ?

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Answer 1 · 2015-12-30T05:55:44

$x = \frac{3}{2}$ $x=3/2$

Explanation:

$ln (x) - ln (x - 1) = ln 3$ $ln(x)-ln(x-1)=ln3$
We know that:

$ln (\frac{y}{z}) = ln y - ln z$ $ln(y/z)=lny-lnz$

Here $y = x and z = x - 1$ $y=x and z=x-1$
$\Rightarrow ln (\frac{x}{x - 1}) = ln 3$ $implies ln(x/(x-1))=ln3$

$\Rightarrow (\frac{x}{x - 1}) = 3$ $implies (x/(x-1))=3$

$\Rightarrow x = 3 (x - 1)$ $implies x=3(x-1)$
$\Rightarrow x = 3 x - 3$ $implies x=3x-3$
$\Rightarrow 2 x = 3$ $implies 2x=3$
$\Rightarrow x = \frac{3}{2}$ $implies x=3/2$

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How do you solve $ln (x) - ln (x - 1) = ln (3)$ $Ln (x) - Ln (x-1) = Ln (3)$ ?

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Explanation:

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How do you solve Ln (x) - Ln (x-1) = Ln (3)ln(x)−ln(x−1)=ln(3)Ln (x) - Ln (x-1) = Ln (3)?

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How do you solve $ln (x) - ln (x - 1) = ln (3)$ $Ln (x) - Ln (x-1) = Ln (3)$ ?