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

Answer (Detailed Solution Below)View answer ▾

Answer: Option B — 125

Detailed Solution

Take $l o g_{5}$ of both sides with $t = l o g_{5} x$ : $(8 t - 24) t = - 4$ .

$8 t^{2} - 24 t + 4 = 0$ , i.e. $2 t^{2} - 6 t + 1 = 0$ (discriminant 28, two real roots).

Product of the $x$ values $= 5^{t_{1}} \cdot 5^{t_{2}} = 5^{t_{1} + t_{2}} = 5^{3} = 125$ .

✓ Check

t_{1} + t_{2} = \frac{6}{2} = 3

from the quadratic.

⚠ Common trap

Using the product of roots

t_{1} t_{2} = \frac{1}{2}

instead of the sum gives

5

; multiplying powers adds exponents.

Quick tip

$x_{1} x_{2} = 5^{t_{1} + t_{2}} = 5^{6/2} = 125$

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Logarithm — Preparation

This question tests your understanding of Logarithm, a key topic in the Quantitative Aptitude section. The detailed solution above uses a step-by-step approach to show the reasoning, and where available, a shortcut that helps you solve similar questions in under 90 seconds during the actual exam.

A logarithm answers the question: to what power must a base be raised to give a number? If a^x = N, then log base a of N = x. This topic covers the laws of logarithms (product, quotient and power rules), change of base, common and natural logarithms, and solving logarithmic and exponential equations. Most exam questions reduce to applying the three main rules correctly: log(mn) = log m + log n, log(m/n) = log m - log n, and log(m^k) = k log m. Remember that logarithms are only defined for positive numbers and a base that is positive and not equal to 1, which helps you reject invalid answers. Practise the logarithm questions below. Each comes with a step-by-step solution showing which law is applied at each step.

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