Question

How many solutions the equation ${\log }_{x^2}16+{\log }_{2x}64=3$ has?

• A. one irrational solution
• B. no solution is a prime number
• C. two real solutions
• D. one integral solution

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A one integral solution
B one irrational solution
C two real solutions
D no solution is a prime number

The equation given is,
${\log }_{x^2}16+{\log }_{2x}64=3$
Equation is defined when,
$x>0$ and $x\ne \frac{1}{2},1$
Now $\frac{1}{2}{\log }_x{4}^2+{\log }_{2x}{4}^3=3$
$\Rightarrow {\log }_{x}4+3\left({\log }_{2x}4\right)=3,[\because {\log }_{a^n}{b}^m=\frac{m}{n}{\log }_{a}b]$
$\Rightarrow 2{\log }_{x}2+6{\log }_{2x}2=3$
$\Rightarrow \frac{2}{\left({\log }_{2}x\right)}+\frac{6}{\left({\log }_{2}x\right)}=3,[\because {\log }_{b}a=\frac{1}{{\log }_{a}b}]$
$\Rightarrow \frac{2}{\left({\log }_{2}x\right)}+\frac{6}{({\log }_{2}2+{\log }_{2}x)}=3$
$\Rightarrow \frac{2}{t}+\frac{6}{(1+t)}=3,[\because \log (ab)=\log a+\log b\text{\&}{\log }_{a}a=1]$, where $t={\log }_{2}x$
$\Rightarrow 2+2t+6t=3t+3t^2$
$\Rightarrow 3t^2-5t-2=0$
$\Rightarrow 3t^2-6t+t-2=0$
$\Rightarrow 3t(t-2)+1(t-2)=0$
$\Rightarrow t=2,-\frac{1}{3}$
$\Rightarrow {\log }_{2}x=2,-\frac{1}{3}$
$\therefore x=4,2^{-1/3}$