Math

QuestionSolve the equation with rational exponents: (x5)2/3=64(x-5)^{2/3} = 64. Select the correct choice: A. The solution set is x=10x = 10 B. The solution set is the empty set

Studdy Solution

STEP 1

Assumptions
1. We are given the equation with a rational exponent: (x5)23=64(x-5)^{\frac{2}{3}} = 64.
2. We need to solve for the variable xx.
3. The exponent 23\frac{2}{3} means the expression is squared and then the cube root is taken, or vice versa.

STEP 2

To eliminate the rational exponent, we can raise both sides of the equation to the reciprocal of 23\frac{2}{3}, which is 32\frac{3}{2}.
((x5)23)32=6432\left((x-5)^{\frac{2}{3}}\right)^{\frac{3}{2}} = 64^{\frac{3}{2}}

STEP 3

When we raise a power to a power, we multiply the exponents. The left side simplifies because 23×32=1\frac{2}{3} \times \frac{3}{2} = 1.
(x5)23×32=(x5)1 (x-5)^{\frac{2}{3} \times \frac{3}{2}} = (x-5)^1

STEP 4

Simplify the left side of the equation.
x5=6432 x-5 = 64^{\frac{3}{2}}

STEP 5

Calculate the right side by first taking the square root of 64 and then cubing the result.
6412=8 64^{\frac{1}{2}} = 8
83=512 8^3 = 512

STEP 6

Substitute the calculated value back into the equation.
x5=512 x-5 = 512

STEP 7

Add 5 to both sides of the equation to solve for xx.
x=512+5 x = 512 + 5

STEP 8

Calculate the value of xx.
x=517 x = 517
The solution set is x=517x = 517.
A. The solution set is {517} \{517\} . B. The solution set is not the empty set.

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