# How do you solve log x + log(x+21) =2?

Jun 20, 2018

$x = 4$

#### Explanation:

Here,

$\log x + \log \left(x + 21\right) = 2$

We assume that the common base of the log as 10.

${\log}_{10} x + {\log}_{10} \left(x + 21\right) = 2$

=>log_10(x*(x+21))=2to[becauselogM+logN=log(MN)

$\implies {\log}_{10} \left({x}^{2} + 21 x\right) = 2$

$\implies {x}^{2} + 21 x = {10}^{2} \to \left[\because X = {\log}_{a} Y \iff Y = {a}^{X}\right]$

$\implies {x}^{2} + 21 x - 100 = 0$

$\implies {x}^{2} + 25 x - 4 x - 100 = 0$

$\implies x \left(x + 25\right) - 4 \left(x + 25\right) = 0$

$\implies \left(x - 4\right) \left(x + 25\right) = 0$

$\implies x = 4 \mathmr{and} x = - 25$

But for $x = - 25$ log x is not defined.

Hence $x = 4$

Check :

$L H S = {\log}_{10} 4 + {\log}_{10} \left(4 + 21\right) = {\log}_{10} 4 + {\log}_{10} 25$

$\therefore L H S = {\log}_{10} \left(4 \times 25\right) = {\log}_{10} 100 = {\log}_{10} {10}^{2} = 2 = R H S$