Question

How do you solve `ln(2x-5) - ln(4x)= 2`?

Answer 1

I found: `x=5/(2(1-2e^2)`. But it cannot be accepted.

Explanation:

I would use one of the rules of logs as:
`log_ax-log_ay=log_a(x/y)` to get:
`ln((2x-5)/(4x))=2` solving the log (with `ln=log_e`) you get:
`(2x-5)/(4x)=e^2`
rearranging:
`2x-5=4xe^2`
`2x-4xe^2=5`
`2x(1-2e^2)=5`
so that:
`x=5/(2(1-2e^2))=-0.363`
BUT it is negative!
If you substitute back the arguments of the `ln` become negative!!!
We cannot accept them; so NO (real) SOLUTIONS!