Question

Find the value of x:

`sqrt(5^0 + 3/7) = (0.49)^(2 - 3x)`

Answer 1

Given expression is `sqrt(5^0 + 3/7) = (0.49)^(2 - 3x)`.

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

Thus, `sqrt(5^0 + 3/7) = (0.49)^(2 - 3x)`

`(5^0 + 3/7)^(1/2) = (49/100)^(2 - 3x)`  ...`[∴ root(n)(a) = a^(1/n)]`

`(1 + 3/7)^(1/2) = (7^2/10^2)^(2 - 3x)`   ...[∴ a⁰ = 1, when a ≠ 0]

`(10/7)^(1/2) = (7/10)^(2(2 - 3x))`   ...[∴ (aⁿ)^m = a^nm]

`(7/10)^((-1)/2) = (7/10)^(4 - 6x)`  ...`[∴ (a/b)^n = (b/a)^-n]`

Equating the powers with same bases.

`(-1)/2 = 4 - 6x`

–1 = 2(4 – 6x)

–1 = 8 – 12x

12x = 8 + 1

12x = 9

⇒ `x = 9/12`

⇒ `x = 3/4`

Therefore, the value of x in the given expression is `3/4`.