Question

For what value of b does (one-twelfth) Superscript negative 2 b Baseline times 12 Superscript negative 2 b + 2 Baseline = 12?

Answer 1

Answer:

Option d. No solution

Step-by-step explanation:

The rules of exponents are:

$1) a^m*a^n = a^{m+n}\\2)a^m/a^n = a^{m-n}\\3) \frac{1}{a} ^m = a^{-m}$

Given:

(one-twelfth) Superscript negative 2 b Baseline times 12 Superscript negative 2 b + 2 Baseline = 12

The algebraic expression for the given problem is:

$(\frac{1}{12} )^{-2b}*(12)^{-2b+2}=12\\(12)^{-(-2b)}*(12)^{-2b+2}=12\\(12)^{2b}*(12)^{-2b+2}=12\\(12)^{2b-2b+2}=12^1$

The base of the left hand side = The base of the right hand side

So, the exponents of the two sides will be equal.

∴ 2b - 2b + 2 = 1

∴ 2 = 1 (rejected because un-logic condition)

So, the answer is d. No solution

Answer 2

Answer:

D. No solution

Step-by-step explanation: