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

Jul 23, 2016

no solution

#### Explanation:

Since $\ln a - \ln b = \ln \left(\frac{a}{b}\right) \mathmr{and} \ln e = 1$

you have the equivalent equation:

$\ln \left(\frac{x}{x + 1}\right) = \ln e$

Then:

$\frac{x}{x + 1} = e$

You also must fix:

$x > 0 \mathmr{and} x + 1 > 0$

that is $x > 0$

and

$x = e \left(x + 1\right)$

$x = e x + e$

$x - e x = e$

$x \left(1 - e\right) = e$

$x = \frac{e}{1 - e}$

but it is a negative number, against the domain of the solution fixed as a positive number, then the equation has no solution