How do you solve  log (x+4)=log x + log 4?

Feb 7, 2016

Put all logs to one side of the equation and solve using the property ${\log}_{a} n - {\log}_{a} m = {\log}_{a} \left(\frac{n}{m}\right)$

Explanation:

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

$\log \left(\frac{\frac{x + 4}{x}}{4}\right) = 0$

Convert to exponential form. The base is 10, since nothing is noted in subscript in the log.

$\left(\frac{\frac{x + 4}{x}}{4}\right) = {10}^{0}$

$\frac{x + 4}{x} = 1 \times 4$

x + 4 = 4x

4 = 3x

$\frac{4}{3}$ = x

Hopefully this helps.

Feb 7, 2016

$x = \frac{4}{3}$

Explanation:

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

$\log \left(\frac{x + 4}{x}\right) = \log \left(4\right)$

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

$\log \left(\frac{x + 4}{x} \times \frac{1}{4}\right) = 0$

$\log \left(\frac{x + 4}{4 x}\right) = 0$

Consider standard form: ${\log}_{10} \left(x\right) = y \to {10}^{y} = x$

so$\text{ } {\log}_{10} \left(\frac{x + 4}{4 x}\right) = 0 \to {10}^{0} = \frac{x + 4}{4 x}$

Giving: $\text{ } 4 x = x + 4$

$3 x = 4$

$x = \frac{4}{3}$