log(2x+1)=1+log(x-2)
I am assuming these are common logarithms, where the base is 10.
log(10) = 1
log(2x+1)=log(10) + log(x-2)
The sum of logs (of the same base) is the log of the product.
log(2x+1) = log[10(x-2)] = log(10x-20)
Remove the log from left and right.
2x+1 = 10x-20
Move all terms with the x to the left side and all other terms to the right side. Remember to switch signs when you switch sides.
2x-10x = -20 - 1
-8x = -21
x = 21/8
Now go back and make sure it works with each of the logs in the original questions. The parameter to a log can not be negative.
log(2x+1) = log(2(21/8)+1) = log(21/4+1) = log(25/4)
The argument is positive, so this is fine.
log(x-2) = log(21/8-2) = log(21/8-16/8) = log(5/8)
This is still positive. So this x=21/8 works for all logs, therefore it is a valid answer.