Natural Log Equation

May 2010
22
0
I have the problem

ln(c)+ln(2-c)=3
I get c=2 or c=0. Then when I plug those values in, it's wrong. Is there even a solution? Is it undefined because of the -c?
Thanks.
 
Feb 2010
1,036
386
Dirty South
I have the problem

ln(c)+ln(2-c)=3
I get c=2 or c=0. Then when I plug those values in, it's wrong. Is there even a solution? Is it undefined because of the -c?
Thanks.
Since \(\displaystyle logA+logB=log(AB)\) you can write

\(\displaystyle log((2-c)c)=3\)

\(\displaystyle c(2-c)=e^3\)

finish it..
 
Nov 2009
517
130
Big Red, NY
I have the problem

ln(c)+ln(2-c)=3
I get c=2 or c=0. Then when I plug those values in, it's wrong. Is there even a solution? Is it undefined because of the -c?
Thanks.
\(\displaystyle \log(c)+\log(2-c)=3\)

\(\displaystyle \implies \log{(c(2-c))}=3\)

\(\displaystyle \implies c(2-c)=e^3\)

\(\displaystyle \implies -c^2+ 2c- e^3 = 0\)

Now, use the quadratic equation to find your solutions.

\(\displaystyle \frac{-2 \pm \sqrt{2^2 - 4 (-1)(- e^3)}}{2(-1)}\)

You may get a complex or two.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
I have the problem

ln(c)+ln(2-c)=3
I get c=2 or c=0. Then when I plug those values in, it's wrong. Is there even a solution? Is it undefined because of the -c?
Thanks.
First, there is no "-c". There is "2- c" which will be positive as long as c< 2. Second, c= 0 can't possibly be a solution specifically because ln is defined only for positive argument.

Others have shown you how to solve that. Just out of curiosity how did you "get c= 2 or c= 0"? Did you set c= 0 and 2- c= 0?