Basics & LogarithmseasyPYQ · JEE Main · 22 Jan 2025 · Shift 1 (Morning)Free

Basics & Logarithms: Product Solutions Equation (JEE Main 2025)

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

Question
The product of all solutions of the equation e5(log⁡ex)2+3=x8e^{5(\log_{e}x)^{2}+3}=x^{8}, x>0x>0, is
Ae2e^{2}
Bee
Ce8/5e^{8/5}correct
De6/5e^{6/5}
Solution
Step 1: Take natural log of both sides:
ln⁡(e5(ln⁡x)2+3)=ln⁡(x8)\ln\left(e^{5(\ln x)^{2}+3}\right)=\ln(x^{8})
5(ln⁡x)2+3=8ln⁡x5(\ln x)^{2}+3=8\ln x
 Step 2: Let t=ln⁡xt=\ln x. Then:
5t2−8t+3=05t^{2}-8t+3=0
 Step 3: Solve by factoring or quadratic formula:
t=8±64−6010=8±210t=\dfrac{8\pm\sqrt{64-60}}{10}=\dfrac{8\pm 2}{10}
 So t1=1t_{1}=1 and t2=35t_{2}=\dfrac{3}{5}. Step 4: By Vieta's formulas:
t1+t2=85=ln⁡x1+ln⁡x2=ln⁡(x1x2)t_{1}+t_{2}=\dfrac{8}{5}=\ln x_{1}+\ln x_{2}=\ln(x_{1}x_{2})
 Step 5: Therefore:
x1x2=e8/5x_{1}x_{2}=e^{8/5}
 Answer: (3) e8/5e^{8/5}
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