# Logarithm Doubling question

• May 22nd 2013, 01:34 PM
Gurp925
Logarithm Doubling question
Hey there,My questions state

Solve algebraically without using logarithms (1/8)^2x = 16^x-5

And : a population of bacteria in a glass of milk on the counter is quadrupling every half an hour
A) write a function for growth p(t) where t is in hours
B) now t in minutes

For a I got p(.5)=A(4)^t(.5) I put .5 because it's in hours so half of one
for B I got p(30)=a(4)^t/30

• May 22nd 2013, 02:20 PM
zhandele
Re: Logarithm Doubling question
I haven't done this kind of thing in a while, but let's see ...

Without using logarithms, convert everything to powers of 2,

$\displaystyle \frac{\1}{8}=2^{-3}\\(2^{-3})^{2x}=2^{-6x}\\16^{x-5}=({2^4})^x({2^4})^{-5}=2^{4x-20}$

You should be able to set the exponents equal and solve for x. As for the others, I'll post again as soon as I can work out the Latex, but if the population quadruples in half an hour, won't it increase 16 fold in a whole hour?
• May 22nd 2013, 02:21 PM
Plato
Re: Logarithm Doubling question
Quote:

Originally Posted by Gurp925
Solve algebraically without using logarithms (1/8)^2x = 16^x-5

That can be written as $\displaystyle 2^{-6x}=2^{4x-20}$
• May 22nd 2013, 02:43 PM
zhandele
Re: Logarithm Doubling question
If $\displaystyle p(t)$ is population as a function of time, and t is in hours, and the population increases 16 fold in an hour, and $\displaystyle p_0$ is the population at time 0, then I'd think

$\displaystyle p(t) = {p_0}16^t$

If $\displaystyle t=0$, what is $\displaystyle p(t)$? If $\displaystyle t=1$, what is $\displaystyle p(t)$? Finally, if $\displaystyle t={\frac{1}{2}$ then what is $\displaystyle p(t)$?

If t is in minutes, then you have to figure out what to put in the place of the 16. Let x be the number we want ....

$\displaystyle x^{60}=16\\60{lnx}=ln{16}\\x=16^{\frac{1}{60}}$

But like I said, I haven't done this in a while. I might be wrong.
• May 22nd 2013, 04:10 PM
Gurp925
Re: Logarithm Doubling question
Im sorry zhandele i really do not understand your second post its a little confusing, is there any way you could simplify your answer?

I do however get my answer for the problem stated by Plato
• May 22nd 2013, 04:50 PM
zhandele
Re: Logarithm Doubling question
I'm not sure where the difficulty is. Could you tell me what isn't clear?

Let's see. The problem doesn't say how many bacteria there are "to start with," and it doesn't give a time when the bacteria start to multiply. But we have to start from some time, with some number of bacteria. Let's call the start time

$\displaystyle t_0$

Are you with me so far? Now let's call the number of bacteria we have "to start with"

$\displaystyle p_0$

Maybe this well help. Suppose

$\displaystyle p_0=100$

This means that the population or number of bacteria at time zero is 100. And we know that a half hour later, we have four times as many, so if we measure time in hours then

$\displaystyle p_{.5}=400$

and if we're measuring time in minutes, then

$\displaystyle p_{30}=400$

After a second half-hour, these numbers would quadruple again. Is that clear? So in hours

$\displaystyle p_{1}=1600$

and in minutes

$\displaystyle p_{60}=1600$

What we're doing is multiplying the initial population, which we don't know, by 16 fold for each hour that passes. The number of hours is the variable t in

$\displaystyle p(t)=p_0 16^t$

By $\displaystyle p(t)$ I mean "p as a function of t."

If we measure t in minutes, then

$\displaystyle p(t)=p_0 16^{\frac{t}{60}}$

Is that better? Is it clear why I divided the t by 60? Substitute in t = 60 (for one hour, 60 minutes) or t = 30 (for one half hour, 30 minutes) and see what happens.

Basically, I'm converting hours into minutes by dividing by 60. I would also take the 60th root of 16, then raise to the t power. That would give an equivalent result, but it would appear as a decimal approximation.