Question

How do you solve `log_4x=log_8(4x)`?

Answer 1

`x = 2^4`

Explanation:

Using

`log_a b= log_e a/(log_e b)`

`log_4x = (log_e x)/(log_e 4) = log_8(4x) = (log_e (4x))/(log_e 8)`

but `log_e 4 = 2 log_e 2` and `log_e 8 = 3 log_e 2` so

`log_e x/(2log_e 2) = (log_e x + 2log_e 2)/(3 log_e 2)`

`3log_e x = 2log_e x+4 log_e 2`

and finally

`log_e x = log_e 2^4`

so

`x = 2^4`