Question

How do you solve `log_2 (x) + log_6 (x) = 3`?

Answer 1

`x = 2^((3(log_2 3+1))/(log_2 3 + 2))`

Explanation:

`log_a b = log_c b/log_c a`
`log (ab) = log a + log b`
`log_a a = 1`
`log_a b = c <=> a^c = b`

`log_2 x + log_6 x = 3`
`log_2 x + log_2 x/log_2 6 = 3`
`log_2 x(1+1/log_2 6) = 3`
`log_2 x = (3log_2 6)/(log_2 6 + 1)`
`x = 2^((3log_2 6)/(log_2 6 + 1))`
`x = 2^((3(log_2 3+1))/(log_2 3 + 2))`