# What is x if log(x+10) -log (3x-8)= 2?

Mar 11, 2018

see a solution step below..

#### Explanation:

$\log \left(x + 10\right) - \log \left(3 x - 8\right) = 2$

Recall the law of logarithm..

$\log a - \log b = \log \frac{a}{\log} b$

Hence;

$\log \frac{x + 10}{\log} \left(3 x - 8\right) = 2 \to \text{applying law of logarithm}$

$\log \left[\frac{x + 10}{3 x - 8}\right] = 2$

For every log, there is a base, and it's default base is $10$, since there was no actual base specified in the given question..

Therefore;

${\log}_{10} \left[\frac{x + 10}{3 x - 8}\right] = 2$

Recall again the law of logarithm..

${\log}_{a} x = y \Rightarrow x = {a}^{y}$

Hence;

$\frac{x + 10}{3 x - 8} = {10}^{2} \to \text{applying law of logarithm}$

$\frac{x + 10}{3 x - 8} = 100$

$\frac{x + 10}{3 x - 8} = \frac{100}{1}$

$1 \left(x + 10\right) = 100 \left(3 x - 8\right) \to \text{cross multiplying}$

$x + 10 = 300 x - 800$

$300 x - 800 = x + 10 \to \text{rearranging the equation}$

$300 x - x = 10 + 800 \to \text{collecting like terms}$

$299 x = 810$

$x = \frac{810}{299} \to \text{diving both sides by the coefficient of x}$

$x = 2.70$