How do I solve this natural logarithm equation?

2 ln(3x - 4) = 7. First, I'd divide both sides by 2.
ln(3x - 4) = 3.5. Now ln both sides.
3x - 4 = ln(3.5). Add 4 to both sides.
3x = ln(3.5) + 4. Divide both sides by 3.
x = (ln(3.5) + 4)/3
x = approximately 1.7509.

Hope this helps.

Hrm... If I remember log, I'd divide the two out so it says:

ln (3x-4) = 3.5 Then you find e^3.5 as the equation essentially now says:

e^3.5 = (3x-4)

Then it's just simple algebra, add 4 to both sides and divide both sides by 3.

You might want to double check your math book though (best place for information) it's been a while since I did log equations.

ln(3x-4) = 7/2
3x-4 = e^(7/2)
x = [ e^(7/2) + 4 ] / 3
x ~= 12.37