Question

If log₂ x = a, log₃ y = a, find `24^(2a + 1)` in terms of x and y.

Answer 1

Given: log₂ x = a, log₃ y = a

Step-wise calculation:

1. From the logarithms:

x = 2^a and y = 3^a

2. `24^(2a + 1) = 24 xx 24^(2a)`

3. Write 24 = 2³ × 3,

So 24^2a = (2³ × 3)^2a

24^2a = 2^6a × 3^2a

4. Therefore `24(2a + 1) = 24 xx 2^{6a} xx 3^{2a}` 

`24^(2a + 1) = 24 xx (2^a)^6 xx (3^a)^2`

`24^(2a + 1) = 24x^6y^2`