Question

Find the value of x:

`root(3)(4^0 + 3/5) = (5/8)^(1 - 2x)`

Answer 1

Given expression is `root(3)(4^0 + 3/5) = (5/8)^(1 - 2x)`.

We need to find the value of x in the given expression.

Thus, `root(3)(4^0 + 3/5) = (5/8)^(1 - 2x)`

`(4^0 + 3/5)^(1/3) = (5/8)^(1 - 2x)`  ...`[∴ root(n)(a) = a^(1/n)]`

`(1 + 3/5)^(1/3) = (5/8)^(1 - 2x)`   ...[∴ a⁰ = 1, when a ≠ 0]

`(8/5)^(1/3) = (5/8)^(1 - 2x)`

`(5/8)^((-1)/3) = (5/8)^(1 - 2x)`  ...`[∴ (a/b)^n = (b/a)^-n]`

Equating the powers with same bases.

`(-1)/3 = 1 - 2x`

–1 = 3(1 – 2x)

–1 = 3 – 6x

6x = 3 + 1

6x = 4

⇒ `x = 4/6`

⇒ `x = 2/3`

Therefore, the value of x in given expression is `2/3`.