`log_2 5x - log_4 5x = 3`

we cannot use quotient rule now, since the terms are of different bases

Hence,

from: `log_b a = (log a)/(log b)`

⇒ `log_2 5x - (log 5x)/(log 2^2) = 3`

⇒ `log_2 5x - (log 5x)/(2log 2) = 3`

⇒ `log_2 5x - 1/2 * (log 5x)/(log 2) = 3`

⇒ `log_2 5x - 1/2 * log_2 5x = 3`

let `log_2 5x = a`

Now,

⇒ `a - a/2 = 3`

⇒ `a/2 = 3` ⇒ `a = 2` x 3 = 6

Hence,

`log_2 5x = a = 6`

⇒ `5x = 2^6`

`5x = 64` ⇒ `(5x)/5 = (64)/5 = 12.8`

∴ The value of `x` is `12.8`

Sorry, it's under construction !

Sorry, it's under construction !