SOLUTION: The problem I have is {{{y = 2e^(3t-1)}}} and i have to make "t" the subject. I have got to {{{t = (ln2 - lny)/3 +1}}}. But i doubt very much that is right.

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: The problem I have is {{{y = 2e^(3t-1)}}} and i have to make "t" the subject. I have got to {{{t = (ln2 - lny)/3 +1}}}. But i doubt very much that is right.       Log On


   

Question 30867: The problem I have is y+=+2e%5E%283t-1%29 and i have to make "t" the subject.
I have got to t+=+%28ln2+-+lny%29%2F3+%2B1. But i doubt very much that is right.
Answer by acerX(62) About Me  (Show Source):
You can put this solution on YOUR website!
y+=+2e%5E%283t-1%29
ln%28y%29+=+ln%282e%5E%283t-1%29%29 introduce the natural log
ln%28y%29=+ln%282%29+%2B+ln%28e%5E%283t-1%29%29 Product Rule of logs
ln%28y%29+=+ln%282%29+%2B+%283t-1%29%2Aln%28e%29 Ln of e is 1 as per definition of ln
ln%28y%29+=+ln%282%29+%2B+3t-1 subtract the ln2 and add the 1 to both sides
ln%28y%29+-+ln%282%29+%2B1+=+3t divide both sides by 3 and use the quotient rule of logs
t+=+%28ln%28y%29+-+ln%282%29+%2B1%29%2F3
So Darryl yes your answer is correct but not in the most simple form