How do you solve $ln x - ln (x + 2) = 1$ $lnx-ln(x+2)=1$ ?

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Answer 1 · 2018-03-20T03:02:53

$x = \frac{2 e}{1 - e}$ $x=(2e)/(1-e)$

First use the property of logarithms that $ln (a) - ln (b) = ln (\frac{a}{b})$ $ln(a)-ln(b)=ln(a/b)$ .

$ln x - ln (x + 2) = ln (\frac{x}{x + 2}) = 1$ $lnx-ln(x+2)=ln(x/(x+2))=1$

Now take the exponential of both sides.

$e^{ln (\frac{x}{x + 2})} = \frac{x}{x + 2}$ $e^ln(x/(x+2))=x/(x+2)$ and $e^{1} = e$ $e^1=e$ so

$\frac{x}{x + 2} = e$ $x/(x+2)=e$

Multiply both sides by $x + 2$ $x+2$

$x = e (x + 2)$ $x=e(x+2)$

Use the distributive property.

$x = e x + 2 e$ $x=ex+2e$

Subtract $e x$ $ex$ from both sides and factor the left hand side.

$x (1 - e) = 2 e$ $x(1-e)=2e$

Divide both sides by $(1 - e)$ $(1-e)$ .

$x = \frac{2 e}{1 - e}$ $x=(2e)/(1-e)$ .

How do you solve lnx−ln(x+2)=1lnx-ln(x+2)=1?