Question

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

• A. $-\frac{3}{2}$
• B. $-\frac{23}{2}$
• C. $5$
• D. both $5$ and $-\frac{23}{2}$

Answer 1

The correct option is C $5$

Clearly, $x-1>0;2x+15>0...\left(1\right)$
${\log }_{10}\sqrt{x-1}+\frac{1}{2}{\log }_{10}(2x+15)=1\Rightarrow {\log }_{10}\sqrt{x-1}+{\log }_{10}\sqrt{2x+15}=1\Rightarrow {\log }_{10}\sqrt{(x-1)(2x+15)}=1\Rightarrow \sqrt{(x-1)(2x+15)}=10\Rightarrow 2x^2+13x-115=0\Rightarrow (x-5)(2x+23)=0\Rightarrow x=5,-\frac{23}{2}\therefore x=5(\because \text{from eqn}(1\left)\right)$