Logarithms PYQ | JEE Main 2025 April | Fastest Trick

 

❓ Question

Given

$$e^2\cdot x^{\ln x}=x^3,$$

where

$$x>\beta,\qquad \alpha^m=\beta^n,$$

and $m,n$ are coprime.

Find the value of

$$m\times n.$$

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✍️ Solution

Given,

$$e^2\cdot x^{\ln x}=x^3.$$

Taking natural logarithm on both sides,

$$\ln\left(e^2\cdot x^{\ln x}\right)=\ln(x^3).$$

Using

$$\ln(ab)=\ln a+\ln b,$$

and

$$\ln(a^b)=b\ln a,$$

we get

$$2+(\ln x)^2=3\ln x.$$

Rearranging,

$$(\ln x)^2-3\ln x+2=0.$$

Let

$$t=\ln x.$$

Then,

$$t^2-3t+2=0.$$

Factorizing,

$$(t-1)(t-2)=0.$$

Hence,

$$t=1\quad\text{or}\quad t=2.$$

Therefore,

$$x=e \quad\text{or}\quad x=e^2.$$

Since

$$\alpha>\beta,$$

we have

$$\boxed{\alpha=e^2,\qquad \beta=e.}$$

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Now,

$$\alpha^m=\beta^n$$

gives

$$(e^2)^m=e^n.$$

Hence,

$$2m=n.$$

Since $m$ and $n$ are coprime,

$$\boxed{m=1,\qquad n=2.}$$

Therefore,

$$m\times n=1\times2=2.$$

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✅ Final Answer

$$\boxed{2}$$