Question

Solve for y:

`root(3)(5^0 + 3/4) = (4/7)^(2 - y)`

Answer 1

We are solving:

`root(3)(5^0 + 3/4) = (4/7)^(2 - y)`

Step 1: Simplify inside cube root

5⁰ = 1 ⇒ `5^0 + 3/4 = 1 + 3/4 = 7/4`

So, `root(3)(7/4) = (4/7)^(2 - y)`

Step 2: Rewrite cube root in exponent form

`root(3)(7/4) = (7/4)^(1/3)`

So equation becomes:

`(7/4)^(1/3) = (4/7)^(2 - y)`

Step 3: Express with same base

`7/4 = (4/7)^-1`

So, `(7/4)^(1/3) = (4/7)^(-1/3)`

Thus, `(4/7)^(-1/3) = (4/7)^(2 - y)`

Step 4: Equating exponents

`-1/3 = 2 - y`

Solve for y:

`y = 2 + 1/3`

= `6/3 + 1/3`

= `7/3`

`y = 2 1/3`