উদাহরণ ৪৷সমাধান করো: 
[H.S’95,Jt Ent’81]
সমাধান:
{\log _x}2 \cdot {\log _{\frac{x}{{16}}}}2 = {\log _{\frac{x}{{64}}}}2\\
\Rightarrow {\log _x}2 \cdot \frac{1}{{{{\log }_2}\frac{x}{{16}}}} = \frac{1}{{{{\log }_2}\frac{x}{{64}}}}\\
\Rightarrow {\log _x}2 \cdot {\log _2}\frac{x}{{64}} = {\log _2}\frac{x}{{16}}\\
\Rightarrow {\log _x}2\left( {{{\log }_2}x - {{\log }_2}64} \right) = {\log _2}x - {\log _2}16\\
\Rightarrow {\log _x}2 \cdot {\log _2}x - {\log _x}2 \cdot {\log _2}{2^6} = {\log _2}x - {\log _2}{2^4}\\
\Rightarrow 1 - {\log _x}2 \cdot 6{\log _2}2 = {\log _2}x - 4{\log _2}2\\
\Rightarrow 1 - 6{\log _x}2 = {\log _2}x - 4\\
\Rightarrow 6{\log _x}2 = 5 - {\log _2}x\\
\Rightarrow 6\frac{1}{{{{\log }_2}x}} = 5 - {\log _2}x\\
\Rightarrow 5{\log _2}x - {\left( {{{\log }_2}x} \right)^2} = 6\\
\Rightarrow {\left( {{{\log }_2}x} \right)^2} - 5{\log _2}x + 6 = 0\\
\Rightarrow {a^2} - 5a + 6 = 0\left[ {{{\log }_2}x = a} \right]\\
\Rightarrow \left( {a - 3} \right)\left( {a - 2} \right) = 0\\
a = 3 \Rightarrow {\log _2}x = 3 \Rightarrow x = {2^3} = 8\\
or,a = 2 \Rightarrow {\log _2}x = 2 \Rightarrow x = {2^2} = 4
\end{array}" alt="\begin{array}{l}
{\log _x}2 \cdot {\log _{\frac{x}{{16}}}}2 = {\log _{\frac{x}{{64}}}}2\\
\Rightarrow {\log _x}2 \cdot \frac{1}{{{{\log }_2}\frac{x}{{16}}}} = \frac{1}{{{{\log }_2}\frac{x}{{64}}}}\\
\Rightarrow {\log _x}2 \cdot {\log _2}\frac{x}{{64}} = {\log _2}\frac{x}{{16}}\\
\Rightarrow {\log _x}2\left( {{{\log }_2}x - {{\log }_2}64} \right) = {\log _2}x - {\log _2}16\\
\Rightarrow {\log _x}2 \cdot {\log _2}x - {\log _x}2 \cdot {\log _2}{2^6} = {\log _2}x - {\log _2}{2^4}\\
\Rightarrow 1 - {\log _x}2 \cdot 6{\log _2}2 = {\log _2}x - 4{\log _2}2\\
\Rightarrow 1 - 6{\log _x}2 = {\log _2}x - 4\\
\Rightarrow 6{\log _x}2 = 5 - {\log _2}x\\
\Rightarrow 6\frac{1}{{{{\log }_2}x}} = 5 - {\log _2}x\\
\Rightarrow 5{\log _2}x - {\left( {{{\log }_2}x} \right)^2} = 6\\
\Rightarrow {\left( {{{\log }_2}x} \right)^2} - 5{\log _2}x + 6 = 0\\
\Rightarrow {a^2} - 5a + 6 = 0\left[ {{{\log }_2}x = a} \right]\\
\Rightarrow \left( {a - 3} \right)\left( {a - 2} \right) = 0\\
a = 3 \Rightarrow {\log _2}x = 3 \Rightarrow x = {2^3} = 8\\
or,a = 2 \Rightarrow {\log _2}x = 2 \Rightarrow x = {2^2} = 4
\end{array}"/>
