NIMCET 2025 Paper

Let us put $t=\log_5 x$. Since $x>0$, this is well-defined, and we can also write $x=5^t$. Now look at the given equation:$x^{8 \log_5 x-24}=5^{-4}$Substituting $\log_5 x=t$ and $x=5^t$, the left-hand side becomes$(5^t)^{8t-24}=5^{t(8t-24)}$So the equation turns into$5^{t(8t-24)}=5^{-4}.$Since the base 5 is the same on both sides and nonzero, we can equate the exponents:$t(8t-24)=-4.$Expanding, we get$8t^2-24t+4=0.$Divide the whole equation by 4 to simplify:$2t^2-6t+1=0$This is a quadratic in $t$. Using the quadratic formula,$t = \frac{6 \pm \sqrt{(-6)^2-4 \cdot 2 \cdot 1}}{2 \cdot 2} = \frac{6 \pm \sqrt{36-8}}{4} = \frac{6 \pm \sqrt{28}}{4} = \frac{6 \pm 2\sqrt{7}}{4} = \frac{3 \pm \sqrt{7}}{2}.$So there are two possible values of $t$ :$t_1 = \frac{3+\sqrt{7}}{2}, \qquad t_2 = \frac{3-\sqrt{7}}{2}.$Recall that $x=5^t$. Therefore, the corresponding values of $x$ areRecall that $x=5^t$. Therefore, the corresponding values of $x$ are$x_1 = 5^{t_1} = 5^{\frac{3+\sqrt{7}}{2}}, \qquad x_2 = 5^{t_2} = 5^{\frac{3-\sqrt{7}}{2}}.$We need the product of all possible values of $x$, that is $x_1 x_2$. Multiply them:$x_1 x_2 = 5^{\frac{3+\sqrt{7}}{2}} \cdot 5^{\frac{3-\sqrt{7}}{2}} = 5^{\frac{(3+\sqrt{7})+(3-\sqrt{7})}{2}} = 5^{\frac{6}{2}} = 5^3 = 125.$So the product of all possible values of $x$ is 125 .