Question

Find the value of x in the following:

`log_2 x + log_8 x + log_32 x = 23/15`

Answer 1

Given: `log_2 x + log_8 x + log_32 x = 23/15`

Step-wise calculation:

1. Let y = log₂ x.

2. Then log₈ x 

= `log_(2^3) x`

= `y/3`

And log₃₂ x 

= `log_(2^5) x`

= `y/5`

3. Substitute:

`y + y/3 + y/5 = 23/15`

4. Combine terms:

`y(1 + 1/3 + 1/5)`

= `y(23/15)`

= `23/15`

5. Therefore y = 1. 

So log₂ x = 1.

⇒ x = 2