If $x < - 1$ and $2^{∣ x + 1 ∣} - 2^{x} = ∣ 2^{x} - 1 ∣ + 1$ , then the value of $x$ is:

Answer (Detailed Solution Below)View answer ▾

Answer: Option A — $- 2$

Detailed Solution

For $x < - 1$ : $x + 1 < 0$ , so $∣ x + 1 ∣ = - (x + 1)$ ; and $0 < 2^{x} < 1$ , so $∣ 2^{x} - 1 ∣ = 1 - 2^{x}$ .

The equation becomes $2^{- x - 1} - 2^{x} = 2 - 2^{x}$ , i.e. $2^{- x - 1} = 2$ .

$- x - 1 = 1$ , so $x = - 2$ .

✓ Check

x = - 2

: left side

= 2^{1} - \frac{1}{4} = \frac{7}{4}

; right side

= \frac{1}{4} - 1 + 1 = \frac{7}{4}

⚠ Common trap

(D) 6 (the source key) breaks the given condition

x < - 1

, and so do (B) and (C).

Quick tip

$x < - 1 : 2^{- x - 1} - 2^{x} = 2 - 2^{x} \Rightarrow 2^{- x - 1} = 2 \Rightarrow x = - 2$

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Indices — Preparation

This question tests your understanding of Indices, 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.

Indices, also called exponents or powers, show how many times a number is multiplied by itself. This topic covers the laws of indices: a^m x a^n = a^(m+n), a^m / a^n = a^(m-n), (a^m)^n = a^(mn), a^0 = 1, negative indices (a^-n = 1/a^n), and fractional indices (a^(1/n) is the nth root of a). It also covers surds and solving exponential equations. The standard approach for exponential equations is to write both sides with the same base and then equate the powers. For example, 4^x = 32 becomes 2^(2x) = 2^5, so x = 5/2. Comparing large powers is often easiest after rewriting them with a common exponent. Practise the indices questions below, each with a step-by-step solution showing which law of indices is used.

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If $x < - 1$ and $2^{∣ x + 1 ∣} - 2^{x} = ∣ 2^{x} - 1 ∣ + 1$ , then the value of $x$ is:

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If x<−1 and 2∣x+1∣−2x=∣2x−1∣+1, then the value of x is:

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If $x < - 1$ and $2^{∣ x + 1 ∣} - 2^{x} = ∣ 2^{x} - 1 ∣ + 1$ , then the value of $x$ is: