Basics & LogarithmseasyJEE Main 2025Free

Basics & Logarithms — JEE Maths practice question (JEE Main 2025)

JEE Maths question with a full step-by-step solution.

Question

The product of all solutions of the equation

e^{5(\log_{e}x)^{2}+3}=x^{8}

x>0

, is

e^{2}

e

e^{8/5}

correct

e^{6/5}

Solution

Step 1: Take natural log of both sides:

\ln\left(e^{5(\ln x)^{2}+3}\right)=\ln(x^{8})

5(\ln x)^{2}+3=8\ln x

Step 2: Let

t=\ln x

. Then:

5t^{2}-8t+3=0

Step 3: Solve by factoring or quadratic formula:

t=\dfrac{8\pm\sqrt{64-60}}{10}=\dfrac{8\pm 2}{10}

t_{1}=1

and

t_{2}=\dfrac{3}{5}

. Step 4: By Vieta's formulas:

t_{1}+t_{2}=\dfrac{8}{5}=\ln x_{1}+\ln x_{2}=\ln(x_{1}x_{2})

Step 5: Therefore:

x_{1}x_{2}=e^{8/5}

Answer: (3)

e^{8/5}

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