How do you solve 2logx = log(-2x+15)?

Aug 9, 2015

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

Explanation:

$2 \log x = \log \left(- 2 x + 15\right)$

Recall that $a \log x = \log \left({x}^{a}\right)$, so

$\log {x}^{2} = \log \left(- 2 x + 15\right)$

Convert the logarithmic equation to an exponential equation.

${10}^{\log {x}^{2}} = {10}^{\log \left(- 2 x + 15\right)}$

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

${x}^{2} = - 2 x + 15$

${x}^{2} + 2 x - 15 = 0$

$\left(x + 5\right) \left(x - 3\right) = 0$

$x + 5 = 0$ and $x - 3 = 0$

$x = - 5$ and $x = 3$

Check:

$2 \log x = \log \left(- 2 x + 15\right)$

If $x = - 5$,

$2 \log \left(- 5\right) = \log \left(- 2 \left(- 5\right) + 15\right)$

This is impossible, because $\log \left(- 5\right)$ is undefined.

If $x = 3$,

2log3= log(-2×3+15)

$\log {3}^{2} = \log \left(- 6 + 15\right)$

$\log 9 = \log 9$

$x = 3$ is a solution.