Question

The solution set of equation log_x 2 log_2x 2 = log_4x 2 is
1. \({2^{-\sqrt{2}}, 2^{\sqrt{2}}}\)
2. \(\left\lbrace \dfrac{1}{2}, 2 \right\rbrace\)
3. \(\left\lbrace \dfrac{1}{4}, 2^2 \right\rbrace\)
4. \(\left\lbrace \dfrac{1}{4}, 2 \right\rbrace\)

Answer 1

Correct Answer - Option 1 : \({2^{-\sqrt{2}}, 2^{\sqrt{2}}}\)

Concept:

​Logarithm properties:

Product Rule	\(\rm \log (mn) =log(m)+log(n)\)
Quotient Rule	\(\rm \log (\frac{m}{n}) =log(m)-log(n)\)
Power Rule	\(\rm \log (m^n) =n\text{ }log(m)\)
Change of Base	\(\boldsymbol {\rm log_m(n)=\frac{1}{log_n(m)}=\frac {log(n)}{log(m)}}\)

 

Calculation:

 logx 2 log2x 2 = log4x 2

\(\rm \because log_x2=\frac{1}{log_2x}\), \(\rm log_{2x}2=\frac{1}{log_22x}\) and \(\rm log_{4x}2=\frac{1}{log_24x}\)

⇒ \(\rm\frac{1}{log_2x}*\frac{1}{log_22x}=\frac{1}{log_24x}\)

⇒ \(\rm{log_2x}\text{ }{log_22x}={log_24x}\)

⇒ \(\rm{log_2x}({log_22}+log_2{x})={log_24}+log_2{x}\)

Let \(\rm{log_2x}\) be a, 

⇒ \(\rm a(1+a)={log_22^2}+a\)      \(\rm(\because log_22 =1)\)

⇒ \(\rm a+a^2=2+a\)

⇒ \(\boldsymbol{\rm a =\pm\sqrt2}\)

⇒ \(\rm log_2x =\pm\sqrt2\)

⇒ \(\boldsymbol{\rm x =2^{\sqrt2},2^{-\sqrt2}}\)