Question

Solve the following equation:
$\frac{1}{2}{\log }_{10}x+3{\log }_{10}\sqrt{2+x}={\log }_{10}\sqrt{x\left(x+2\right)}+2$

Answer 1

$\frac{1}{2}{\log }_{10}x+3{\log }_{10}\sqrt{2+x}={\log }_{10}\sqrt{x(x+2)}+2$

$\Rightarrow \frac{1}{2}{\log }_{10}x+\frac{3}{2}{\log }_{10}(2+x)=\frac{1}{2}{\log }_{10}\left(x\right(x+2\left)\right)+2\dots ({\log }_y{x}^n=n{\log }_{y}x)$

$\Rightarrow {\log }_{10}x+3{\log }_{10}(2+x)={\log }_{10}x(x+2)+4$

$\Rightarrow {\log }_{10}x(x+2{)}^3-{\log }_{10}\left(x\right(x+2\left)\right)=4\dots (\log ab=\log a+\log b)$

$\Rightarrow {\log }_{10}\left[\frac{x(x+2{)}^3}{x(x+2)}\right]=4\dots (\log a-\log b=\log \frac{a}{b})$

$\Rightarrow {\log }_{10}(x+2{)}^2=4$

$\Rightarrow (x+2{)}^2={10}^4$

$\Rightarrow (x+2{)}^2-{10}^4=0$

$\Rightarrow (x+2-{10}^2)(x+2+{10}^1)=0\dots (a^2-b^2=(a+b\left)\right(a-b\left)\right)$

$\Rightarrow (x-98)(x+102)=0$

$\Rightarrow x=98,-102$