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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
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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