How can I use logarithms to determine the growth rate of a bacteria colony?

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We're on the section of logarithms, so I'm guessing this has something to do with them...

If the size of a bacteria colony doubles in 5 hours, how long will it take for the number of bacteria to triple?

I know that doubling time and all that stuff is independent of the amount of something you have...

Should I use P = P0at or P = P0ekt ?

I'll just try it both ways to show some work, but I don't know which equation i'd be using...

first with P = P0at
2 = 1a5
2 = a5
a = 21/5
a = 1.1487

3 = (1.1487)t
ln3 = tln(1.1487)
t = ln3 / ln1.1487
t = 7.92473 hours


Now, to try using P = P0ekt

2 = 1e5k
2 = e5k
ln2 = 5k
k = ln2 / 5
k = .138629

3 = 1e.138629t
3 = e.138629t
ln3 = .138629t
t = ln3 / .138629
t = 7.92484 hours

very close answers, which one equation should i have used, and is the answer right?
 
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A quick glance should convince you that ##a=e^k##, so the answers are equivalent.
 
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Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...

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