exponentials/logarithms/compound interest

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amberlianne

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Dec 7, 2006
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Whenever I try and solve an exponential equation using logarithms, I get stuck at the point where I'm dividing one log by the other, ending up with a larger number in the denominator than the numerator, and stupid results with compound interest problems (solving for T) like 0.002345 years or some such, obviously incorrect.

EX:

Find out how long it takes an investment of $1500 to triple if it is invested at 8% interest and compounded monthly.

4500=1500(1+(0.08/12))^12t
log4500=12tlog1500(1+0.08/12))
log4500/12log1510.00000000001=t
3.65321251377534/38.147723367518=t
t=0.0957648895211916

As this is obviously not correct, and I've been trying to figure this out using the book and other online resources for over a week now, I really need some explanation as to what I'm doing wrong.

Thanks.
 
Let's see.

\(\displaystyle \L\\4500=1500(1+\frac{\frac{2}{25}}{12})^{12t}\)

\(\displaystyle \L\\3=(1+\frac{\frac{2}{25}}{12})^{12t}\)

\(\displaystyle \L\\ln(3)=12tln(1+\frac{\frac{2}{25}}{12})\)

\(\displaystyle \L\\ln(3)=12tln(\frac{151}{150})\)

\(\displaystyle \L\\\frac{ln(3)}{12ln(\frac{151}{150})}=t\)
 
amberlianne said:
Whenever I try and solve an exponential equation using logarithms, I get stuck at the point where I'm dividing one log by the other, ending up with a larger number in the denominator than the numerator, and stupid results with compound interest problems (solving for T) like 0.002345 years or some such, obviously incorrect.

EX:

Find out how long it takes an investment of $1500 to triple if it is invested at 8% interest and compounded monthly.

4500=1500(1+(0.08/12))^12t
log4500=12tlog1500(1+0.08/12))
log4500/12log1510.00000000001=t
3.65321251377534/38.147723367518=t
t=0.0957648895211916

As this is obviously not correct, and I've been trying to figure this out using the book and other online resources for over a week now, I really need some explanation as to what I'm doing wrong.

Thanks.
Click to expand...

Do some simplifying BEFORE you take the log of both sides.....then you're less likely to make the mistakes you've made.

Divide both sides of the equation by 1500:

3 = (1 + (.08/12))<SUP>12t</SUP>

Now, take the log of both sides:

log 3 = log (1 + (.08/12))<SUP>12t</SUP>

log 3 = 12t log (1 + (.08/12))

Divide both sides by 12 log(1 + (.08/12))

log 3 / 12 log(1 + (.08/12)) = t

Be careful entering the expression on the left into your calculator....I'd use ALL of the parentheses!

I got t = 13.778

I hope this helps you.
 
Is there any way to do this without a calculator which has the parentheses function?
 
Hello, amberlianne!

You're taking logs too early . . .

Find out how long it takes an investment of $1500 to triple
if it is invested at 8% interest and compounded monthly.
Click to expand...

You started off fine . . .

\(\displaystyle 1500\left(1\,+\,\frac{0.08}{12}\right)^{12t}\;=\;4500 \;\;\Rightarrow\;\; \left(1\,+\,\frac{0.08}{12}\right)^{12t} \;=\;3\)

Now take logs: \(\displaystyle \:\ln\left(1\,+\,\frac{0.08}{12}\right)^{12t} \;=\;\ln3\)

. . and we have: \(\displaystyle \:12t\,\cdot\,\ln\left(1\,+\,\frac{0.08}{12}\right) \;=\;\ln(3)\)


Therefore: \(\displaystyle t \;=\;\frac{\ln(3)}{12\left(1\,+\,\frac{0.08}{12}\right)} \;=\;13.77837843 \:\approx\:14\text{ years}\)

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Do you have a calculator which has \(\displaystyle \fbox{\log}\) and \(\displaystyle \fbox{\ln}\) but no \(\displaystyle \fbox{(}\) or \(\displaystyle \fbox{)}\) ??

 
Yes, I do. And it's only a free-download calculator anyways. I'm hoping my brother or someone will let me borrow a real one for the test. Funny how I made it until a week before finals with nothing but a 99-cent-store solar calculator, and now all of the sudden I need this crazy expensive one. Ah well. I suppose I'll just have to download one with the parentheses function for homework purposes.

Thanks, all.
 
I got a better calculator and I still can't do this stuff right. Are there any really helpful websites out there that can explain exponentials and logarithms to me in a manner which will finally make it make sense?
 
I don't see why you're having problems, or that you need an exotic calculator:

we have: 12t = log(3) / log(1 + .08/12)
12t = log(3) / log(1.006666.....)

Get log(1.006666...) and store result in a memory location :
you must have at least one memory location on your cheap calculator!;
if not, just write the darn thing down)

Now get log(3): divide result by what's in memory;
divide that result by 12, and bingo!

log(1.006666...) = .00664454.....
log(3) = 1.09861228....

1.09861228 / .00664454 = 165.3405413....

165.3405413 / 12 = 13.778378...

Note: if you were trying with Soroban's, notice that he forgot a "log"
in the denominator...his 1st mistake this year :wink:
 
Haha, yeah, I pretended that never happened and inserted it anyways when I tried with his example.

I really don't know what the problem is. I understand the theory just fine. Then I go and try and work a problem step-by-step along with the "show me an example" function of the online homework, and they divide something and I divide the same thing but we end up with different answers. Even when I work off of all y'all's examples, when I divide something you've divided, the answer is different. I have downloaded five different free calculators now, thinking maybe the answer is in this cheap free software, but I don't know where to go really, now. Somehow I have to scrape an 80% on the homework or I won't be allowed to take the test.

Anyways, thanks for all the help and such.
 
amberlianne said:
I have downloaded five different free calculators now, thinking maybe the answer is in this cheap free software, but I don't know where to go really, now.
Click to expand...
Amber, go to http://www.google.com and enter this in search box:
log(3) / log(1 + (.08 / 12)) / 12

Sooooooooo......... :shock:
 
Sorry for being rude.... I completely forgot to thank you all for your help with this! Update: I took some of what each person's reply was and finally figured this stuff out, and totally aced the course :)

Thanks again!
 
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