# How do you solve  10 log _ 10 (x+21) +log _ 10 (x)= 2?

Sep 2, 2016

$x \approx {10}^{2} / {21}^{10}$

#### Explanation:

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

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

Supposing $x = \delta > 0$ but very small, we will have

${\left(\delta + 21\right)}^{10} \approx {21}^{10}$ so ${21}^{10} \delta \approx {10}^{2}$

$\delta \approx {10}^{2} / {21}^{10}$ then the real solution is

${x}_{0} \approx {10}^{2} / {21}^{10}$