Question

If `2^(3x) = (root(3)(32))^(4/y) = (sqrt(8))^5`, then find the values of x and y.

Answer 1

Given: `2^(3x) = (root(3)(32))^(4/y) = (sqrt(8))^5`

Stepwise calculation:

1. Express each term as a power of 2.

2^3x remains as is.

`root(3)(32) = 32^(1//3)`

Since 32 = 2⁵, this becomes (2⁵)^1/3 = 2^5/3.

`(root(3)(32))^(4//y) = (2^(5//3))^(4//y)`

`(root(3)(32))^(4//y) = 2^((5//3)*(4//y))`

`(root(3)(32))^(4//y) = 2^(20//(3y))`

`sqrt(8) = 8^(1//2)`

Since 8 = 2³, 

`sqrt(8) = (2^3)^(1/2)`

`sqrt(8) = 2^(3/2)`

`(sqrt(8))^5 = (2^(3//2))^5`

`(sqrt(8))^5 = 2^((3//2)*5)`

`(sqrt(8))^5 = 2^(15//2)`

2. Since all are equal, their exponent powers of 2 must be equal:

`3x = 20/(3y)`

`3x = 15/2`

3. From `3x = 15/2`, 

Solve for (x): 

`x = 15/(2 xx 3)`

`x = 15/6`

`x = 5/2`

4. From `3x = (20)/(3y)`,

Substitute `x = 5/2`:

`3 xx 5/2 = 20/(3y)`

`15/2 = 20/(3y)`

Cross-multiplying:

15 × 3y = 2 × 20 

45y = 40 

`y = 40/45`

`y = 8/9`