Question

If log₂ x = a, log₅ y = a, find `20^(2a  -  1)` in terms of x and y.

Answer 1

Given: log₂ x = a, log₅ y = a

Step-wise calculation:

1. From definitions:

2^a = x and 5^a = y

2. Write 20 = 2² × 5

So `20^(2a - 1) = (2^2 xx 5)^(2a - 1)`

`20^{2a-1} = 2^{4a - 2} xx 5^{2a-1}`

3. Express powers in x and y:

`2^(4a - 2) = 2^(-2) xx (2^a)^4`

`2^(4a - 2) = (1/4) x^4`

`5^(2a - 1) = 5^(-1) xx (5^a)^2`

`5^(2a - 1) = (1/5) y^2`

4. Multiply:

`20^(2a - 1) = (1/4 x^4) xx (1/5 y^2)`

`20^(2a - 1) = (1/20) x^4y^2`

`20^(2a - 1) = (x^4y^2)/20`