উদাহরণ ৮৷ সমাধান করো  - {\log _5}6 - 1 - \frac{1}{{2x}}.jpg "{\log _5}\left( {{5^{\frac{1}{x}}} + 125} \right) = {\log _5}6 + 1 + \frac{1}{{2x}}")
[Jt. Ent.’88]
সমাধান:
{\log _5}\left( {{5^{\frac{1}{x}}} + 125} \right) = {\log _5}6 + 1 + \frac{1}{{2x}}\\
\Rightarrow {5^{\frac{1}{x}}} + 125 = {5^{{{\log }_5}6 + 1 + \frac{1}{{2x}}}}\\
\Rightarrow {5^{\frac{1}{x}}} + 125 = {5^{{{\log }_5}6}} \times 5 \times {5^{\frac{1}{{2x}}}}\\
\Rightarrow {\left( {{5^{\frac{1}{{2x}}}}} \right)^2} + 125 = 6 \times 5 \times \left( {{5^{{\textstyle{1 \over {2x}}}}}} \right)\\
\Rightarrow {a^2} + 125 = 30a,\left[ {{5^{\frac{1}{{2x}}}} = a} \right]\\
\Rightarrow {a^2} - 30a + 125 = 0\\
\Rightarrow {a^2} - \left( {25 + 5} \right)a + 125 = 0\\
\Rightarrow \left( {a - 25} \right)\left( {a - 5} \right) = 0\\
\Rightarrow a = 25,or,a = 5
\end{array}" alt="\begin{array}{l}
{\log _5}\left( {{5^{\frac{1}{x}}} + 125} \right) = {\log _5}6 + 1 + \frac{1}{{2x}}\\
\Rightarrow {5^{\frac{1}{x}}} + 125 = {5^{{{\log }_5}6 + 1 + \frac{1}{{2x}}}}\\
\Rightarrow {5^{\frac{1}{x}}} + 125 = {5^{{{\log }_5}6}} \times 5 \times {5^{\frac{1}{{2x}}}}\\
\Rightarrow {\left( {{5^{\frac{1}{{2x}}}}} \right)^2} + 125 = 6 \times 5 \times \left( {{5^{{\textstyle{1 \over {2x}}}}}} \right)\\
\Rightarrow {a^2} + 125 = 30a,\left[ {{5^{\frac{1}{{2x}}}} = a} \right]\\
\Rightarrow {a^2} - 30a + 125 = 0\\
\Rightarrow {a^2} - \left( {25 + 5} \right)a + 125 = 0\\
\Rightarrow \left( {a - 25} \right)\left( {a - 5} \right) = 0\\
\Rightarrow a = 25,or,a = 5
\end{array}"/>
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