Question

How do you solve `log_4(3x-2)-log_4(4x+1)=2`?

Answer 1

`x = -18/61`
This answer can be verified by substituting in the original equation

Explanation:

The first thing we need to do is to write 2 as a `log_4` term as well.
With logs, the terms must either be all as logs, or all as numbers, not a combination.

`2 = log_4 16, " because " 4^2 = 16`

`log_4(3x-2) - log_4(4x+1) = log_4 16`

If we are subtracting the logs, we must have been dividing the numbers. It is possible to condense two log terms into one using the log laws. `logA - logB = log(A/B)`

`log_4((3x-2)/(4x+1))= log_4 16`

Now because there is one term on each side, the terms which we are finding logs of, must be equal to each other. Drop the logs.

`(3x-2)/(4x+1)= 16 " solve as usual"`

`16(4x +1) = 3x -2`

`64x + 16 = 3x-2`

`61x = -18`

`x = -18/61`