# How do you solve log(2x+1)+log(x-1) = log (4x-4)?

Aug 1, 2015

$\textcolor{red}{x = \frac{3}{2}}$

#### Explanation:

$\log \left(2 x + 1\right) + \log \left(x - 1\right) = \log \left(4 x - 4\right)$

Recall that $\log x + \log y = \log \left(x y\right)$, so

$\log \left(2 x + 1\right) \left(x - 1\right) = \log \left(4 x - 4\right)$

Convert the logarithmic equation to an exponential equation.

${10}^{\log \left(2 x + 1\right) \left(x - 1\right)} = {10}^{\log \left(4 x - 4\right)}$

Remember that ${10}^{\log} x = x$, so

$\left(2 x + 1\right) \left(x - 1\right) = 4 x - 4$

Factor the right hand side.

$\left(2 x + 1\right) \left(x - 1\right) = 4 \left(x - 1\right)$

Divide both sides by $x - 1$, where x≠1.

$2 x + 1 = 4$

$2 x = 3$

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

Check:

$\log \left(2 x + 1\right) + \log \left(x - 1\right) = \log \left(4 x - 4\right)$

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

$\log \left(2 \left(\frac{3}{2}\right) + 1\right) + \log \left(\frac{3}{2} - 1\right) = \log \left(4 \left(\frac{3}{2}\right) - 4\right)$

$\log \left(3 + 1\right) + \log \left(\frac{1}{2}\right) = \log \left(6 - 4\right)$

log4–log1-log2 = log 2

$\log \left(\frac{4}{2}\right) - 0 = \log 2$

$\log 2 = \log 2$

$x = \frac{3}{2}$ is a solution.