The product of all solutions of the equation e^5( e x)^2 + 3 = x^… — Solution | JEE Mathematics

MCQMediumJEE 2025Quadratic Equations in Complex Numbers

The product of all solutions of the equation $e^{5(\log_e x)^2 + 3} = x^8, x > 0$ , is :

Answer

Correct answer:A

Standard Method

Given: $e^{5(\log_e x)^2 + 3} = x^8$ with $x > 0$ .

Find: The product of all solutions.

Take natural logarithm on both sides:

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

So,

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

Let $\ln x = t$ . Then,

5t^2 - 8t + 3 = 0

For the quadratic equation, if the roots are $t_1$ and $t_2$ , then

t_1 + t_2 = \frac{8}{5}

Since $t = \ln x$ , we have

\ln x_1 + \ln x_2 = \frac{8}{5}

Therefore,

\ln(x_1 x_2) = \frac{8}{5}

Hence,

x_1 x_2 = e^{8/5}

Therefore, the product of all solutions is $e^{8/5}$ . The correct option is A.

Root Interpretation

Using $t = \ln x$ converts the given exponential equation into a quadratic in $t$ :

5t^2 - 8t + 3 = 0

The required quantity is not $t_1 t_2$ , but $x_1 x_2$ .

Because $t_1 = \ln x_1$ and $t_2 = \ln x_2$ ,

\ln x_1 + \ln x_2 = \ln(x_1 x_2)

Thus we need the sum of roots, not the product of roots.

From the quadratic,

t_1 + t_2 = \frac{8}{5}

Hence,

\ln(x_1 x_2) = \frac{8}{5}

and so

x_1 x_2 = e^{8/5}

Using $t_1 t_2 = \frac{3}{5}$ to find $x_1 x_2$ is incorrect because $t = \ln x$ . The product of the solutions depends on $\ln x_1 + \ln x_2$ , not on $\ln x_1 \ln x_2$ . Use $\ln(x_1 x_2) = \ln x_1 + \ln x_2$ instead.
Writing $\ln x^8 = (\ln x)^8$ is wrong. The correct logarithmic property is $\ln(x^8) = 8\ln x$ .
Ignoring the condition $x > 0$ is incorrect because $\ln x$ is defined only for positive real $x$ . Keep the domain restriction before substituting $t = \ln x$ .

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