# How do you solve log ( x +1 ) + log 7 = log 14 - log ( 2-x ) ?

Jun 12, 2018

$x = 1$

#### Explanation:

$\log \left(x + 1\right) + \log 7 = \log 14 - \log \left(2 - x\right)$

$\log \left(x + 1\right) + \log \left(2 - x\right) = \log 14 - \log 7$

$\log \left(\frac{x + 1}{2 - x}\right) = \log 2$

$\frac{x + 1}{2 - x} = 2$

$x + 1 = 4 - 2 x$

$3 x = 3$

$x = 1$

Recall:

${\log}_{10} a + {\log}_{10} b = {\log}_{10} \left(a \times b\right) = {\log}_{10} a b$

${\log}_{10} a - {\log}_{10} b = {\log}_{10} \left(a \div i \mathrm{de} b\right) = {\log}_{10} \left(\frac{a}{b}\right)$