Solve for x:
e^4x-3 = 2e^2x1
ln2x+6) - ln(3x-1) = ln(2)
Then, simply the radicals
How about some parentheses?
Just guessing here, I read
e^(4x-3) = 2e^(2x)
No idea what the final 1 is. Anyway, I'll work this, and then you can fix it if necessary.
Taking logs of both sides, we get
4x-3 = ln2 + 2x
2x = 3+ln2
x = (3+ln2)/2
ln(2x+6) - ln(3x-1) = ln(2)
(2x+6)(3x-1) = 2
Now just solve the quadratic