SOLUTION: P=13500+22ln(t+1), where t is th etime in years from the present. In how many years will there be 18,000 omsects?(round to the nearest tenth of a year). 18,000=13,500+22ln(t+1)

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: P=13500+22ln(t+1), where t is th etime in years from the present. In how many years will there be 18,000 omsects?(round to the nearest tenth of a year). 18,000=13,500+22ln(t+1)       Log On


   

Question 240616: P=13500+22ln(t+1), where t is th etime in years from the present. In how many years will there be 18,000 omsects?(round to the nearest tenth of a year).
18,000=13,500+22ln(t+1)
4,500=22ln(t+1)
204.5=ln(t+1)
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
You are correct so far.

18,000=13,500+22ln(t+1)
4,500=22ln(t+1)
204.5=ln(t+1)

The next step is to "undo" the ln function. You can do this by raising both sides as a power of "e".

e%5E204.5=+e%5E%28ln%28t%2B1%29%29
e%5E204.5=+t%2B1

Now, subtract 1 from each side:
t=e%5E204.5-1, which is approximately 6.5046*10^88 years, which is one HECK of a long time!! Are you sure you copied this problem correctly?? Are all the units given in years???

Dr. Robert J. Rapalje, Retired
Seminole State College of Florida
Altamonte Springs Campus