Population growth logarithm

**mattty** · January 14th 2009, 05:06 AM

The question is: The bacterial population in a certain culture is given by $P=I(1.6)^t$ where $t$ is in hours and $I$ is the initial population (when $t=0$ ). Determine how long it takes for this population to double in size.

So, from this, i've worked out that $P=2I$ and then i took a stab at rearranging the equation into the form $x={log_by}$ and got
$t=I\log_{1.6}2I$

Now, although I believe this to be the right answer something isn't ringing true since natural logarithms didn't come into it at all. I've always been told that population growth was one of the main things natural logs were for outside of calculus.

Any help or suggestions would be greatly appreciated.

**galactus** · January 14th 2009, 05:33 AM

Don't over complicate it.

Since it doubles, set P=2I. That way, the I's cancel and you have:

$2=(\frac{8}{5})^{t}$

Now, take log of both sides:

$ln(2)=t\cdot ln(\frac{8}{5})$

$t=\frac{ln(2)}{ln(\frac{8}{5})}$

**HallsofIvy** · January 14th 2009, 05:33 AM

Originally Posted by mattty

The question is: The bacterial population in a certain culture is given by $P=I(1.6)^t$ where $t$ is in hours and $I$ is the initial population (when $t=0$ ). Determine how long it takes for this population to double in size.

So, from this, i've worked out that $P=2I$ and then i took a stab at rearranging the equation into the form $x={log_by}$ and got
$t=I\log_{1.6}2I$

Your equation is, then, $2I= I(1.6)^t$ . The very first thing you should see is that the "I"s cancel, giving [tex]2= (1.6)^t[/quote] so that your answer does not depend on I.

Taking the log, base 1.56 would give $t= log_{1.6}(2)$ but, for some reason, they forgot to put a " $log_{1.6}$ " key on my calculator! Fortunately, logarithms are pretty much interchangable. What do you get if you take ln of both sides? That is, $ln(2)= ln(1.6^t)$ . Use a "law of logarithms" to reduce the right side.

Now, although I believe this to be the right answer something isn't ringing true since natural logarithms didn't come into it at all. I've always been told that population growth was one of the main things natural logs were for outside of calculus.

Any help or suggestions would be greatly appreciated.

**Mush** · January 14th 2009, 05:34 AM

Originally Posted by mattty

The question is: The bacterial population in a certain culture is given by $P=I(1.6)^t$ where $t$ is in hours and $I$ is the initial population (when $t=0$ ). Determine how long it takes for this population to double in size.

So, from this, i've worked out that $P=2I$ and then i took a stab at rearranging the equation into the form $x={log_by}$ and got
$t=I\log_{1.6}2I$

Now, although I believe this to be the right answer something isn't ringing true since natural logarithms didn't come into it at all. I've always been told that population growth was one of the main things natural logs were for outside of calculus.

Any help or suggestions would be greatly appreciated.

You can you logarithms to ANY base for this problem. The reason for using logarithms to solve this problem is that the laws of logarithms allow us to linearise the equation so it can be solved using simple algebra. These laws of logarithms hold for logarithms of ANY base. So whether you use $\ln{}$ or $\log_{2453465345552352345}{}$ doesn't really matter. Choose to use natural logarithms if you please:

$P = 2I$

$\rightarrow I(1.6)^t= 2I$

$\rightarrow (1.6)^t= 2$

$\rightarrow \ln{(1.6)^t}= \ln{2}$

$\rightarrow t \ln{(1.6)}= \ln{2}$

$\therefore t = \frac{\ln{2}}{\ln{(1.6)}}$

**mattty** · January 14th 2009, 06:18 AM

Originally Posted by galactus

Don't over complicate it.

It's very much my worst trait, i always tend to overcomplicate everything unfortunately.

Originally Posted by HallsofIvy

Your equation is, then, $2I= I(1.6)^t$ . The very first thing you should see is that the "I"s cancel, giving [itex]2= (1.6)^t[/quote ] so that your answer does not depend on I.

Taking the log, base 1.56 would give $t= log_{1.6}(2)$ but, for some reason, they forgot to put a " $log_{1.6}$ " key on my calculator!

I guess I was just too eager to convert to the [math x={log_by}[/tex] form and didn't notice, feeling a bit foolish now

on the upside though, at least my calculator does come with a " $log_{1.6}$ " key, gotta love the ti nspires

many thanks to all 3 of you anyway

**galactus** · January 14th 2009, 06:29 AM

Wow, the TI inspire has a built-in log base change?. That's cool.

On all calculators I have seen, even my beloved Voyage 200, one still has to change bases if one wants a base other than 10.

At least, I think so. Maybe I don't know what I'm doing.

**mattty** · January 14th 2009, 06:47 AM

yep, it doesn't even default to base 10, when you press the log button you automatically need to put in the base and x values yourself. the nspire is great for working with algebra too, if only i could figure out how to set domains and ranges for the graphs i would be estatic -_-

**Mush** · January 14th 2009, 07:30 AM

Originally Posted by mattty

yep, it doesn't even default to base 10, when you press the log button you automatically need to put in the base and x values yourself. the nspire is great for working with algebra too, if only i could figure out how to set domains and ranges for the graphs i would be estatic -_-

My Casio fx-570ES does that. It's not big deal!

**mr fantastic** · January 14th 2009, 12:28 PM

Originally Posted by galactus

Wow, the TI inspire has a built-in log base change?. That's cool.

On all calculators I have seen, even my beloved Voyage 200, one still has to change bases if one wants a base other than 10.

At least, I think so. Maybe I don't know what I'm doing.

Depending on what OS version you're running, the TI-89 does it too.

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