Question

Sometimes to solve an equation, we may use the identity $a^{{\log }_{a}b}=b,b>0,a>0,a\ne 1$.

then solution set of $x^{{\log }_{4}x}=2^{3({\log }_{4}x+3)}$ is,

• A. $64,\frac{1}{8}$
• B. $64$
• C. $\frac{1}{8}$
• D. $\varphi$

Answer 1

The correct option is A $64,\frac{1}{8}$

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

Above equation is valid when $x>0$

Now taking log both sides on base 2 of given equation,

$\left({\log }_{4}x\right)\left({\log }_{2}x\right)=({\log }_{4}x+3){\log }_2\left(8\right)$

Put ${\log }_{2}x=t$ to get

$\Rightarrow \frac{1}{2}{t}^2=(\frac{1}{2}t+3)\left(3\right)[\because {\log }_{2}8={\log }_2{2}^3=3{\log }_{2}2=3]$

$\Rightarrow t^2=3t+18\Rightarrow t=6,-3$

$\Rightarrow x=2^6,2^{-3}=64,\frac{1}{8}$
Ans: A