for what value of b does $\\left(\\frac{1}{12}\\right)^{-2b}\\cdot12^{-2b + 2}=12$?\n- $-\\frac{1}{2}$\n…

for what value of b does $\\left(\\frac{1}{12}\\right)^{-2b}\\cdot12^{-2b + 2}=12$?\n- $-\\frac{1}{2}$\n- $-\\frac{1}{4}$\n- $\\frac{1}{4}$\n- no solution

for what value of b does $\\left(\\frac{1}{12}\\right)^{-2b}\\cdot12^{-2b + 2}=12$?\n- $-\\frac{1}{2}$\n- $-\\frac{1}{4}$\n- $\\frac{1}{4}$\n- no solution

Answer

Explanation:

Step1: Rewrite the first - term

We know that $\left(\frac{1}{12}\right)^{-2b}=(12^{-1})^{-2b}=12^{2b}$ according to the power - of - a - power rule $(a^{m})^{n}=a^{mn}$.

Step2: Combine the terms on the left - hand side

The left - hand side of the equation $\left(\frac{1}{12}\right)^{-2b}\cdot12^{-2b + 2}$ becomes $12^{2b}\cdot12^{-2b + 2}$. According to the product rule of exponents $a^{m}\cdot a^{n}=a^{m + n}$, we have $12^{2b+( - 2b + 2)}=12^{2b-2b + 2}=12^{2}$.

Step3: Solve for $b$

The original equation $12^{2b}\cdot12^{-2b + 2}=12$ becomes $12^{2}=12$, which is a false statement. So the equation has no solution.

Answer:

no solution