Animated Solution for Mathematics - Quadratic Equations: The product of all solutions of the equation $e^{5({\log }_{e}x{)}^2+3}=x^8,x>0$, is :

Imagine you are standing before the equation $e^{5(\ln x{)}^2+3}=x^8$. At first glance, it looks like a fortress where the variable $x$ is locked away in both the base and the exponent.

To unlock this, we apply the natural logarithm, $\ln$, to both sides:

Using the identity $\ln (e^A)=A$, the left side simplifies, and using the power rule $\ln (x^n)=n\ln x$, the right side transforms. We are left with the following equation:

Notice the repetition of the term $\ln x$. This is a classic signal to use substitution.

Let $t=\ln x$. By making this change, the equation morphs into a familiar quadratic form:

We have successfully moved from the world of transcendental functions into the world of parabolas. This quadratic equation represents a curve that intersects the horizontal axis at two points, $t_1$ and $t_2$.

Sum of Roots

We could use the quadratic formula, but there is a more efficient path. The question asks for the product of all solutions for $x$.

If $t_1=\ln x_1$ and $t_2=\ln x_2$, then the product of the solutions is $x_1{x}_2$. We utilize the logarithmic property:

We do not need the individual values of $t_1$ and $t_2$; we only need their sum. From the theory of quadratic equations $a{t}^2+bt+c=0$, the sum of the roots is given by:

Substituting our values $a=5$ and $b=-8$:

We have established that $\ln x_1+\ln x_2=\frac{8}{5}$. By the product rule of logarithms, this simplifies to:

To isolate the product $x_1{x}_2$, we convert the expression back to exponential form. The base $e$ swings over to yield the final result:

$$x_1{x}_2=e^{8/5}$$

This is a beautiful conclusion to a journey that began with a seemingly impenetrable equation. In JEE Advanced, the path is rarely about brute force; it is about finding the most elegant way to reveal the structure hidden beneath the surface.