Aspire Faculty ID #11992 · Topic: NIMCET 2025 · Just now
NIMCET 2025

Let $x$ be a positive real number such that $x^{(8\log_5x-24)}=5^{-4}$. Then the product of all possible values of x is =

Solution

We have $ x^{8\log_5 x - 24} = 5^{-4} $ 
Let $ \log_5 x = t $. 
Then $ x = 5^t $. 
Substitute: $ (5^t)^{8t - 24} = 5^{-4} $ 
$ 5^{t(8t - 24)} = 5^{-4} $ 
So: $ t(8t - 24) = -4 $ 
$ 8t^2 - 24t + 4 = 0 $ 
Divide by 4: 
$ 2t^2 - 6t + 1 = 0 $ 
Solutions: $ t = \frac{3 \pm \sqrt{7}}{2} $ 
Thus: $ x_1 = 5^{\frac{3 + \sqrt{7}}{2}},\ x_2 = 5^{\frac{3 - \sqrt{7}}{2}} $ 
Product: $ x_1 x_2 = 5^{\frac{3 + \sqrt{7}}{2} + \frac{3 - \sqrt{7}}{2}} = 5^{\frac{6}{2}} = 5^3 = 125 $
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