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

Mar 10, 2018

Use the law

$\log \left(a\right) - \log \left(b\right) = \log \left(\frac{a}{b}\right)$.

Hence

$\log \left(\frac{x + 1}{x - 1}\right) = 1$

$\frac{x + 1}{x - 1} = 10$

$x + 1 = 10 \left(x - 1\right)$

$x + 1 = 10 x - 10$

$11 = 9 x$

$x = \frac{11}{9}$

We now verify that it satisfies the equation. It will as long as $x > 1$, which it is. Hence, we're done.

Hopefully this helps!

Mar 10, 2018

The solution is $x = \frac{11}{9}$.

#### Explanation:

Use these $\log$arithm rules to help simplify the equation:

$\log \left(\textcolor{red}{x}\right) \textcolor{g r e e n}{-} \log \left(\textcolor{b l u e}{y}\right) = \log \left(\textcolor{g r e e n}{\frac{\textcolor{red}{x}}{\textcolor{b l u e}{y}}}\right)$

$\log \left(x\right) = \log \left(y\right) q \quad \quad \implies q \quad \quad x = y$

Now here's the actual problem:

$\log \left(\textcolor{red}{x + 1}\right) \textcolor{g r e e n}{-} \log \left(\textcolor{b l u e}{x - 1}\right) = 1$

$\log \left(\textcolor{g r e e n}{\frac{\textcolor{red}{x + 1}}{\textcolor{b l u e}{x - 1}}}\right) = 1$

$\log \left(\textcolor{g r e e n}{\frac{\textcolor{red}{x + 1}}{\textcolor{b l u e}{x - 1}}}\right) = \log \left(10\right)$

$\textcolor{g r e e n}{\frac{\textcolor{red}{x + 1}}{\textcolor{b l u e}{x - 1}}} = 10$

$\textcolor{red}{x + 1} = 10 \left(\textcolor{b l u e}{x - 1}\right)$

$\textcolor{red}{x + 1} = 10 x - 10$

$\textcolor{red}{x} + 11 = 10 x$

$11 = 9 x$

$\frac{11}{9} = x$

$x = \frac{11}{9}$

That's the solution. Hope this helped!