SOLUTION: I have a problem here where I need to use natural log. I am okay doing this, but they threw in an extra variable which makes me unsure of how to proceed. 2e^(3x) + e^(3x) = b

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: I have a problem here where I need to use natural log. I am okay doing this, but they threw in an extra variable which makes me unsure of how to proceed. 2e^(3x) + e^(3x) = b       Log On


   



Question 379855: I have a problem here where I need to use natural log. I am okay doing this, but they threw in an extra variable which makes me unsure of how to proceed.
2e^(3x) + e^(3x) = b , where b is a constant.
I don't understand what to do with b, in order to solve for x?

Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
Just treat b as it were any other constant, such as 10 or 26.3.
We have
2e%5E%283x%29+%2B+e%5E%283x%29+=+b
3e%5E%283x%29+=+b
e%5E%283x%29+=+b%2F3
3x+=+ln+%28b%2F3%29
x+=+%281%2F3%29%28ln%28b%2F3%29%29+=+%281%2F3%29%28ln+b+-+ln+3%29