# How do you solve log(5x+2)=log(2x-5)?

Dec 9, 2015

$x = - \frac{7}{3}$

#### Explanation:

Given $\log \left(5 x + 2\right) = \log \left(2 x - 5\right)$ common log- base 10

Step 1: Raised it to exponent using the base 10

${10}^{\log 5 x + 2} = {10}^{\log 2 x - 5}$

Step 2: Simplify, since ${10}^{\log} A = A$
$5 x + 2 = 2 x - 5$

Step 3: Subtract $\textcolor{red}{2}$ and $\textcolor{b l u e}{2 x}$ to both side of the equation to get
$5 x + 2 \textcolor{red}{- 2} \textcolor{b l u e}{- 2 x} = 2 x \textcolor{b l u e}{- 2 x} - 5 \textcolor{red}{- 2}$
$3 x = - 7$

Step 4: Dive both side by 3
$\frac{3 x}{3} = - \frac{7}{3} \Leftrightarrow x = - \frac{7}{3}$

Step 5: Check the solution

$\log \left[\left(5 \cdot - \frac{7}{3}\right) + 2\right] = \log \left[\left(2 \cdot - \frac{7}{3}\right) - 5\right]$
$\log \left(- \frac{35}{3} + \frac{6}{3}\right) = \log \left(- \frac{14}{3} - \frac{15}{3}\right)$
$\log \left(- \frac{29}{3}\right) = \log \left(- \frac{29}{3}\right)$

Both side are equal, despite we can't take a log of a negative number due to domain restriction ${\log}_{b} x = y , , x > 0 , b > 0$
$x = - \frac{7}{3}$ , assuming a complex valued logarithm