A calculus text I have, by Hoffman and Bradley, poses the following question: “Two plants grow in such a way that t days after planting, they are P1(t) and P2(t) centimeters tall, respectively, where:
P1(t) = 21 / (1+25e^-.03t) and P2(t) = 20/(1+17e^-0.6t)
…at what time do the plants have the same height?"
(There were other questions pertaining to growth rates, and the increase or decrease of these rates, i.e. involving 1st and 2nd derivatives, which were straightforward, and I was able to answer.)
The question regarding the same height, I assume meant setting P1(t) = P2(t), and solving.
It’s easy enough to do on a graphing calculator, which the student solutions manual suggests, and Desmos gave me two points of intersection: -1.118, which is to be discarded, as t is for time, and 20.711 (days), which was the textbook’s answer.
However, before looking to the solutions manual for an answer, I tried solving this algebraically, using natural logarithms, but failed miserably.
There is probably something about rational equations involving exponentials that I’m not understanding at all. Possibly someone could help me out. This was my process:
21/(1+25e^-.03t) = 20/(1+17e^-0.6t)
Cross-multiplying:
21(1+17e^-0.6t) = 20(1 +25e^-0.3t)
21 + 357e^-0.6t = 20 + 500e^-0.3t
Rearranging:
(500e^-0.3t) – (357e^-0.6t) = 1
ln (500e^-0.3t) – ln (357e^-0.6t) = ln 1
(ln 500 + ln e^-0.3t) – (ln 357 + ln e^-0.6t) = ln 1
ln 500 - 0.3t – ln 357 + 0.6t = 0
0.3t = - ln 500 + ln 357
0.3t = -.3368
my answer, t = -.1123, is completely wrong!
Additionally, the graphing calculator gave me two solutions, one negative (to be discarded) and the positive, correct one, of 20.711.
Possibly my calculation was off, but more probably, I made a fundamental error in understanding; maybe both.
What did I do wrong?
Thank You!
You are fine up to this point.
$500e^{-0.3t} – 357e^{-0.6t} = 1$
Your next step is invalid. You cannot say
$\ln\left(500e^{-0.3t} – 357e^{-0.6t}\right) = \ln 1\\ \ln(500e^{-0.3t}) – \ln (357e^{-0.6t}) = 0$
That would be a variation of the freshman's dream.
Anyway what can you do?
$e^{-0.6t} = (e^{-0.3t})^2$
What you have is a quadratic and you can apply the quadratic formula.
Let $u = e^{-0.3t}$
$357u^2 - 500 u + 1 = 0$
$u = \frac {250 + \sqrt {250^2 - 357}}{357}\\ e^{-0.3t}= \frac {250 + \sqrt {250^2 - 357}}{357}$
Note that $e^{-0.3t}>0$ for all $t,$ allowing us to reject
$u = \frac {250 - \sqrt {250^2 - 357}}{357}$ when applying the quadratic formula.
Take our logarithms and simplify.