# simplification of a logarithm

• Feb 24th 2010, 02:43 PM
simplification of a logarithm
Original problem is $\displaystyle 2e^{3x+5}=6$. Ive gotten it to $\displaystyle ln e^{3x+5}=ln3$. Ordinarily I would change the 3x+5 into the coefficient of lne and go from there but in the example problem Im looking at they simply remove the ln e. Basically this is what they did: They took $\displaystyle ln e^{3x+5}=ln3$ and simplified it to $\displaystyle 3x+5=ln 3$. So my question is what rule or property did they use to get rid of the LN e?
• Feb 24th 2010, 03:00 PM
NOX Andrew
They combined two rules. First, they converted the exponent to a coefficient, like you were going to do. Then, they used the rule $\displaystyle \ln{e} = 1$. Multiplication by 1 doesn't change anything so you are left with just the coefficient.
• Feb 24th 2010, 03:56 PM
satx
Quote:

Original problem is $\displaystyle 2e^{3x+5}=6$. Ive gotten it to $\displaystyle ln e^{3x+5}=ln3$. Ordinarily I would change the 3x+5 into the coefficient of lne and go from there but in the example problem Im looking at they simply remove the ln e. Basically this is what they did: They took $\displaystyle ln e^{3x+5}=ln3$ and simplified it to $\displaystyle 3x+5=ln 3$. So my question is what rule or property did they use to get rid of the LN e?