NIMCET 2025 Paper

Let us put $\mathit{t}={\log }_5\mathit{x}$. Since $\mathit{x}>0$, this is well-defined, and we can also write $\mathit{x}=5^{\mathit{t}}$. Now look at the given equation:
${\mathit{x}}^{8{\log }_5\mathit{x}-24}=5^{-4}$
Substituting ${\log }_5\mathit{x}=\mathit{t}$ and $\mathit{x}=5^{\mathit{t}}$, the left-hand side becomes
${\left(5^{\mathit{t}}\right)}^{8\mathit{t}-24}=5^{\mathit{t}\left(8\mathit{t}-24\right)}$
So the equation turns into
$5^{\mathit{t}\left(8\mathit{t}-24\right)}=5^{-4}.$
Since the base 5 is the same on both sides and nonzero, we can equate the exponents:
$\mathit{t}\left(8\mathit{t}-24\right)=-4.$
Expanding, we get
$8{\mathit{t}}^2-24\mathit{t}+4=0.$
Divide the whole equation by 4 to simplify:
$2{\mathit{t}}^2-6\mathit{t}+1=0$
This is a quadratic in $\mathit{t}$. Using the quadratic formula,
$\mathit{t}=\frac{6\pm \sqrt{{\left(-6\right)}^2-4\cdot 2\cdot 1}}{2\cdot 2}=\frac{6\pm \sqrt{36-8}}{4}=\frac{6\pm \sqrt{28}}{4}=\frac{6\pm 2\sqrt{7}}{4}=\frac{3\pm \sqrt{7}}{2}.$
So there are two possible values of $\mathit{t}$ :
${\mathit{t}}_1=\frac{3+\sqrt{7}}{2},{\mathit{t}}_2=\frac{3-\sqrt{7}}{2}.$
Recall that $\mathit{x}=5^{\mathit{t}}$. Therefore, the corresponding values of $\mathit{x}$ are
Recall that $\mathit{x}=5^{\mathit{t}}$. Therefore, the corresponding values of $\mathit{x}$ are
${\mathit{x}}_1=5^{{\mathit{t}}_1}=5^{\frac{3+\sqrt{7}}{2}},{\mathit{x}}_2=5^{{\mathit{t}}_2}=5^{\frac{3-\sqrt{7}}{2}}.$
We need the product of all possible values of $\mathit{x}$, that is ${\mathit{x}}_1{\mathit{x}}_2$. Multiply them:
${\mathit{x}}_1{\mathit{x}}_2=5^{\frac{3+\sqrt{7}}{2}}\cdot 5^{\frac{3-\sqrt{7}}{2}}=5^{\left(\frac{3+\sqrt{7}+3-\sqrt{7}}{2}\right)}=5^{\frac{6}{2}}=5^3=125.$
So the product of all possible values of $\mathit{x}$ is 125 .