Answer:
Option d. No solution
Step-by-step explanation:
The rules of exponents are:
[tex]1) a^m*a^n = a^{m+n}\\2)a^m/a^n = a^{m-n}\\3) \frac{1}{a} ^m = a^{-m}[/tex]
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:
[tex](\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[/tex]
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
