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JEE MainMathematicsToughLogarithm and its applications

The number of solutions of the equation
xlog⁡10x+74=10log⁡10x+1x^{\frac{\log_{10}x + 7}{4}} = 10^{\log_{10}x + 1} is

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xlog⁡10x+74=10log⁡10x+1x^{\frac{\log_{10}x + 7}{4}} = 10^{\log_{10}x + 1}
Take log_10 on both sides
⇒(log⁡10x+74)log⁡10\Rightarrow \left( \frac{\log_{10}x + 7}{4} \right)\log_{10}
x=log⁡1x+1x = \log_{1}x + 1
0\ \ \ \ = 1
Put log_10 u = 1
⇒\Rightarrow t2+3t−4=0t^{2} + 3t - 4 = 0
⇒\Rightarrow t = 1 or t = - 4
⇒\Rightarrow log_10⁡x = 1 or log_10⁡x = - 4
⇒\Rightarrow x = 10, 10^-4

Source: Grass-G / M-05 Mathematics Paper 10 October.docx

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