Question

The number of solutions of $\log \left(2x\right)=2\log (4x-15)$ is

• A. $1$
• B. $2$
• C. $3$
• D. infinite

Answer 1

The correct option is A $1$

$\log \left(2x\right)=2\log (4x-15)$
Above equation is valid when $x>0$
and $4x-15>0\Rightarrow x>\frac{15}{4}$
Now,
$\log \left(2x\right)=2\log (4x-15)=\log (4x-15{)}^2[\because x\log a=\log a^x]$
$\Rightarrow 2x={(4x-15)}^2$
$\Rightarrow 16x^2-122x+225=0$
$\Rightarrow x=\frac{9}{2}$
Ans: A