Question

The set \( \{ x \in \mathbb{R} : 16(2^x)>16^{x-1} \} \) is:

• A. \( \{ x \in \mathbb{R} : x>0 \} \)
• B. \( \{ x \in \mathbb{R} : x < 0 \} \)
• C. \( \mathbb{R} \)
• D. \( \{ x \in \mathbb{R} : x>2 \} \)

Hint:

When solving inequalities with exponents, simplify the terms and compare the powers to find the solution.

Answer 1

The Correct Option is A

We are given the inequality \( 16(2^x)>16^{x-1} \). Let's simplify it: \[ 16(2^x) = 16 \cdot 2^x = 2^{4} \cdot 2^x = 2^{x+4} \] \[ 16^{x-1} = (2^4)^{x-1} = 2^{4(x-1)} = 2^{4x-4} \] Thus, the inequality becomes: \[ 2^{x+4}>2^{4x-4} \] Comparing the exponents: \[ x + 4>4x - 4 \] Solving for \( x \): \[ 4 + 4>4x - x \] \[ 8>3x \quad \Rightarrow \quad x < \frac{8}{3} \] Thus, the solution set is \( \{ x \in \mathbb{R} : x>0 \} \).