Solve: [(1/7)^-1]^4 Using Laws of Negative Exponents

Question

[(17)1]4= [(\frac{1}{7})^{-1}]^4=

Video Solution

Solution Steps

00:00 Solve
00:03 Negative exponent switches between numerator and denominator
00:09 And this is the solution to the question

Step-by-Step Solution

We use the power property of a negative exponent:

an=1an a^{-n}=\frac{1}{a^n} We will rewrite the fraction in parentheses as a negative power:

17=71 \frac{1}{7}=7^{-1} Let's return to the problem, where we had:

((17)1)4=((71)1)4 \bigg( \big( \frac{1}{7}\big)^{-1}\bigg)^4=\big((7^{-1})^{-1} \big)^4 We continue and use the power property of an exponent raised to another exponent:

(am)n=amn (a^m)^n=a^{m\cdot n} And we apply it in the problem:

((71)1)4=(711)4=(71)4=714=74 \big((7^{-1})^{-1} \big)^4 =(7^{-1\cdot-1})^4=(7^1)^4=7^{1\cdot4}=7^4 Therefore, the correct answer is option c

Answer

74 7^4