Using Log to find interest rate: $5K to $7.5K in 5 years w/ interest compounded monthly

[doakland]

Thanks in advance. We are supposed to use logarithms to answer this question.

Question: If interest is compounded monthly, at what annual rate must you invest $5,000 to have $7,500 at the end of 5 years?

I understand it up to the point where I have it simplified, but then I don't know how to use logarithm to solve. Thank you!!!

I just noticed an error - 1.25 should be 1.5. The question still remains - how do I use a log to solve the equation? Thank you!

 

[skeeter]

two more errors ... 60 monthly compounding periods in 5 years

[imath]7500 = 5000\left(1+\dfrac{r}{12}\right)^{60}[/imath]

[imath]1.5 = \left(1+\dfrac{r}{12} \right)^{60}[/imath]

[imath]\log(1.5) = 60\log\left(1+\dfrac{r}{12}\right)[/imath]

[imath]\dfrac{\log(1.5)}{60} = \log\left(1+\dfrac{r}{12}\right)[/imath]

[imath]b^{\frac{\log(1.5)}{60}} = 1 + \dfrac{r}{12}[/imath], where [imath]b[/imath] is whatever log base you wish to use

can you finish?

 

[Steven G]

What does n represent? How many month are in a year?
Why are you being asked to solve using logs? Can we please see the exact problem?

If I were was asked to use logs here is how I would proceed. 1st note that if logA = logB (for any common base), then A=B.

As skeeter said you have 1.5 = (1 + r/12)⁶⁰.
Now to live up to using logs I would then write:
log1.5 = log(1 + r/12)⁶⁰ (so I am using logs)
Then I would get (again!) that 1.5 = (1 + r/12)⁶⁰
You know what (1 + r/12)⁶⁰ equals (it equals 1.5), so what 1 + r/12 equal.


I doubt that you were asked to use logs. I am sure that this problem came from the section on logs and compound interest, but not every one of these problems need to be solve using logs.

 

[lookagain]

We are supposed to use logarithms to answer this question.

No, this is not a problem to use logarithms. The variable is part of the base.

\(\displaystyle 1.5 = \bigg(1 + \dfrac{r}{12}\bigg)^{60}\)

\(\displaystyle (1.5)^{\tfrac{1}{60}} = 1 + \dfrac{r}{12}\)

Please continue from here.

 

[NotJeffM]

As has been pointed out, if you have a good calculator, you do not NEED to use logarithms, but you can.

Historically, before modern calculators, you WOULD have needed to use logs (to base 10 most likely) as follows.

[math]1.5 = \left ( 1 + \dfrac{r}{12} \right )^{60} \implies 1.5^{1/60} = 1 + \dfrac{r}{12}.\\ z = (1.5)^{1/60} \implies \log(z) = \dfrac{\log(1.5)}{60} \approx 0.002934854 \implies \\ z \approx \log^{-1}(0.002934854) = 1.0067806 \implies \\ 1 + \dfrac{r}{12} \approx 1.0067806 \implies r \approx 12 * 0.0067806 = 8.137\%. [/math]
But in the modern world you simply go

[math] 1 + \dfrac{r}{12} = 1.5^{1/60} \approx 1.0067806 \implies r \approx 12 * 0.00687 = 8.137\%. [/math]
The calculator now does the work that used to be done with log tables and pencil and paper.