Question

Find the value of x in the following:

`log_3 x + log_9 x + log_81 x = 7/4`

Answer 1

Given: `log_3 x + log_9 x + log_81 x = 7/4`

Step-wise calculation:

1. Write all logs in base 3:

`log_9 x = (log_3 x)/(log_3 9)`

= `(log_3 x)/2`

log₈₁ x

= `(log_3 x)/(log_3 81)`

= `(log_3 x)/4`

2. Substitute into the equation:

`log_3 x + (log_3 x)/2 + (log_3 x)/4 = 7/4`

3. Factor log₃ x: 

`(1 + 1/2 + 1/4) xx log_3 x`

= `7/4 (7/4) xx log_3 x`

= `7/4`

4. Divide both sides by `7/4`: 

log₃ x = 1

⇒ x = 3¹

⇒ x = 3