Question

The number of solution to the equation ${\log }_{10}\sqrt{x-1}+\frac{1}{2}{\log }_{10}(2x+15)=1$ is

Answer 1

${\log }_{10}\sqrt{x-1}+\frac{1}{2}{\log }_{10}(2x+15)=1$
For the $\log$ to be defined,
$x-1>0;2x+15>0\Rightarrow x>1;x>-\frac{15}{2}\Rightarrow x>1\cdots \left(1\right)$
Now,
${\log }_{10}\sqrt{x-1}+\frac{1}{2}{\log }_{10}(2x+15)=1\Rightarrow \frac{1}{2}{\log }_{10}(2x+15)={\log }_{10}10-{\log }_{10}\sqrt{x-1}\Rightarrow {\log }_{10}(2x+15)=2{\log }_{10}\left(\frac{10}{\sqrt{x-1}}\right)\Rightarrow {\log }_{10}(2x+15)={\log }_{10}\left(\frac{10}{\sqrt{x-1}}\right)+{\log }_{10}\left(\frac{10}{\sqrt{x-1}}\right)\Rightarrow {\log }_{10}(2x+15)={\log }_{10}{\left(\frac{10}{\sqrt{x-1}}\right)}^2\Rightarrow 2x+15={\left(\frac{10}{\sqrt{x-1}}\right)}^2\Rightarrow 2x+15=\frac{100}{x-1}\Rightarrow (x-1)(2x+15)=100\Rightarrow 2x^2+13x-115=0\Rightarrow (x-5)(2x+23)=0\Rightarrow x=5,-\frac{23}{2}\therefore x=5(\because x>1)$