# How do you solve Log (x-1) + Log 2 = Log (3x)?

Apr 10, 2018

There is no solution for $x$.

#### Explanation:

By the laws of logarithms, $\log a + \log b = \log \left(a \cdot b\right)$.

Using this law:
$\log \left(x - 1\right) + \log 2 = \log \left(3 x\right)$
$\log \left(2 \left(x - 1\right)\right) = \log \left(3 x\right)$

Using the distributive property ($a \cdot \left(b + c\right) = a b + a c$):
$\log \left(2 x - 2\right) = \log \left(3 x\right)$

${10}^{\log \left(2 x - 2\right)} = {10}^{\log} \left(3 x\right)$
$2 x - 2 = 3 x$
$3 x = 2 x - 2$
Subtracting $2 x$ from both sides:
$x = - 2$

Since you can not apply $\log$ on a negative number, there is no solution for this equation.