# Natural Log Equation

#### dwatkins741

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.

#### harish21

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..

#### Anonymous1

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.
$$\displaystyle \log(c)+\log(2-c)=3$$

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

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

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

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

You may get a complex or two.

#### HallsofIvy

MHF Helper
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?