How do you solve #log_7x-log_7(x-1)=1#?
When working with logs, the terms must all be in either log form or as numbers, but not a combination of both.
In order to have all the terms written as logs with the same base, the given equation can be changed to
If log terms are added, they can be written as the log of one number:
But if log A = log B, then A = B
So we now have: