Evaluate. ((1)/(3))^(-(2)/(5))*96^(-(2)/(5))=

Question

Evaluate.  ((1)/(3))^(-(2)/(5))*96^(-(2)/(5))=

Bytelearn · Accepted Answer

Evaluate first term: Evaluate the first term ((1)/(3))^(-(2)/(5)). To evaluate a negative exponent, take the reciprocal of the base and change the sign of the exponent. ((1)/(3))^(-(2)/(5)) = (3/1)^(2/5)Evaluate second term: Evaluate the second term 96^(-(2)/(5)). First, we need to find the reciprocal of 96 and then raise it to the power of 2/5. 96^(-(2)/(5)) = (1/96)^(2/5)Simplify second term: Simplify the second term (1/96)^(2/5). To simplify a fractional exponent, we can take the fifth root of the base and then square it. (1/96)^(2/5) = (1^(2/5))/(96^(2/5)) Since 1 raised to any power is 1, we have: (1^(2/5))/(96^(2/5)) = 1/(96^(2/5))Find fifth root: Find the fifth root of 96 and then square it. The fifth root of 96 is not a whole number, but we can simplify it by factoring 96 into its prime factors. 96 = 2^5 * 3 The fifth root of 96 is the fifth root of (2^5 * 3), which is 2 * fifth root of 3. Now, square this result: (2 * fifth root of 3)^2Combine results: Combine the results from Step 1 and Step 4. We have (3/1)^(2/5) for the first term and 1/(2 * fifth root of 3)^2 for the second term. Now, multiply these two expressions: (3/1)^(2/5) * 1/(2 * fifth root of 3)^2Simplify multiplication: Simplify the multiplication. Since (3/1)^(2/5) is the same as the fifth root of 3 squared, we can write it as (fifth root of 3)^2. (fifth root of 3)^2 * 1/(2 * fifth root of 3)^2 Now, we can cancel out (fifth root of 3)^2 from the numerator and denominator. The result is 1/2^2.Calculate final result: Calculate the final result. 1/2^2 = 1/4

Evaluate. $\newline$ $\left(\frac{1}{3}\right)^{-\frac{2}{5}} \cdot 96^{-\frac{2}{5}}=$

Full solution

More problems from Powers with negative bases

Related topics

Evaluate.\newline(13)−25⋅96−25= \left(\frac{1}{3}\right)^{-\frac{2}{5}} \cdot 96^{-\frac{2}{5}}= (31​)−52​⋅96−52​=

Evaluate. $\newline$ $\left(\frac{1}{3}\right)^{-\frac{2}{5}} \cdot 96^{-\frac{2}{5}}=$