Find the value of x, if (2^1log_x4)(2^1log_x16)(2^1log_x256)⋯=2.

A. 2. Step 1: 1log_x4=log_4x=log_2x2; likewise 1log_x16=log_2x4 and 1log_x256=log_2x8, and so on. Step 2: The product is 2^log_2x(1/2+1/4+1/8+⋯)=2^log_2x=x. Step 3: So x=2. Check: For x=2: log_24=2, log_216=4, log_2256=8, …, so the product is 2^1/2⋅2^1/4⋅2^1/8⋯=2^1=2. Trap: (B) 1/2 (the source key): log_1/24=-2, so the product becomes 2^-1/2-1/4-⋯=1/2, not 2.

Question 6 of 15 · Logarithm

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NIMCET 2009

Find the value of $x$ , if $(2^{\frac{1}{l o g _{x} 4}}) (2^{\frac{1}{l o g _{x} 16}}) (2^{\frac{1}{l o g _{x} 256}}) \dots = 2$ .

Answer (Detailed Solution Below)View answer ▾

Answer: Option A — $2$

Detailed Solution

$\frac{1}{l o g _{x} 4} = l o g_{4} x = \frac{l o g _{2} x}{2}$ ; likewise $\frac{1}{l o g _{x} 16} = \frac{l o g _{2} x}{4}$ and $\frac{1}{l o g _{x} 256} = \frac{l o g _{2} x}{8}$ , and so on.

The product is $2^{l o g_{2} x (\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots)} = 2^{l o g_{2} x} = x$ .

So $x = 2$ .

✓ Check

For

x = 2

l o g_{2} 4 = 2

l o g_{2} 16 = 4

l o g_{2} 256 = 8

, …, so the product is

2^{1/2} \cdot 2^{1/4} \cdot 2^{1/8} \dots = 2^{1} = 2

⚠ Common trap

(B)

\frac{1}{2}

(the source key):

l o g_{1/2} 4 = - 2

, so the product becomes

2^{- 1/2 - 1/4 - \dots} = \frac{1}{2}

, not 2.

Quick tip

$\frac{1}{l o g _{x} 2 ^{m}} = \frac{l o g _{2} x}{m} \Rightarrow 2^{l o g_{2} x (\frac{1}{2} + \frac{1}{4} + \dots)} = x = 2$

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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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Find the value of $x$ , if $(2^{\frac{1}{l o g _{x} 4}}) (2^{\frac{1}{l o g _{x} 16}}) (2^{\frac{1}{l o g _{x} 256}}) \dots = 2$ .

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Find the value of $x$ , if $(2^{\frac{1}{l o g _{x} 4}}) (2^{\frac{1}{l o g _{x} 16}}) (2^{\frac{1}{l o g _{x} 256}}) \dots = 2$ .