Solve for t. 62e^.07t = (280e^.02t)/2. I know I could just graph them both, but I was hoping there was a symbolic way to compute it. My math toolbox is limited to a little bit of integration (currently in Calc 2), so if your explanation would keep that in mind I'd appreciate it.

Question

Solve for t.  62e^.07t >= (280e^.02t)/2.  I know I could just graph them both, but I was hoping there was a symbolic way to compute it.  My math toolbox is limited to a little bit of integration (currently in Calc 2), so if your explanation would keep that in mind I'd appreciate it.

anonymous · Answer

Can't this be solved with algebra?

whpalmer4 · Answer

$$62e^{0.07t} \ge \frac{280e^{0.02t}}{2}$$Is that the correct formula?

anonymous · Answer

yes, that is it

whpalmer4 · Answer

I would just take the natural log of both sides, and take advantage of $$\ln x^n = n \ln x$$and $$\ln e = 1$$

anonymous · Answer

$$
62e^{.07t} >= (280e^{.02t})/2 \
124e^{.07t} >= 280e^{.02t} \
62e^{.07t} >= (280e^{.02t})/2 \
\ln(124)+.07t >= \ln(280)+.02t \
.05t >= \ln(280)-\ln(124)\
$$

anonymous · Answer

Omg, I'm really embarrassed that I forgot all about ln(xy) = ln(x) + ln(y).  Thank you so much.

whpalmer4 · Answer

Another way to roll would be to divide both sides by $e^{0.02t}$ leaving you with
$$62e^{0.05t} \ge 140$$$$e^{0.05t} \ge\frac{140}{62}$$$$0.05t\ge \ln \frac{140}{62}$$

whpalmer4 · Answer

@wio you can get $\ge\le$ symbols via \ge and \le

anonymous · Answer

I like method #2 as well. Thanks.

whpalmer4 · Answer

What's the context of the problem, out of curiosity?

anonymous · Answer

I was looking ahead in my programming hw and one of the questions was to write a java application that would solve it.  I'm pretty sure the exercise was just to get us to use the math library and just generate discrete values for "t", comparing the output of each function and breaking the while-loop once I had the condition it asked for.  But I got curious about solving it by hand, and got stuck, so I asked here.

whpalmer4 · Answer

Ah, okay, sounds plausible.  I'll caution that in real life, you rarely get to test if a = b when doing mathematical programming :-)

anonymous · Answer

A better paraphrase of the question is:  The population of Mexico in 2009 is 62 mil and has a constant growth rate of 7% per year.  The pop of US is 280 mil with a growth rate of 7% per year.  Assuming the growth rates remain constant, at what year will the pop of Mexico be greater than half that of the US.

anonymous · Answer

er, typo.  US growth rate should be 2%

whpalmer4 · Answer

Of course, the more interesting question is to ask the same thing about the population in California :-)