Math Question...

[mathwhizwannabe]

The Problem:

a. How many years will it be required for $1 invested at 8% to double in value? (Hint: Guess at values of n to find where 1.08 to the nth power is closest to 2.00.)

b. How many years will be required for $100 invested at 8% to double in value?

For a. I followed the hint and figured out the answer to be 9 years. Since a. and b. are both part of the same problem, I assumed there must be some algebraic way to come up with the answer to b. like there was in a. I have worked and worked on this problem (b.), but the only way I got the answer to it was to multiply 100 by .08, then add the answer to 100 and so on until I came up with the number of years it would require. Is there a more algebraic way to figure this out? The book doesn't go into much detail. (For b. my answer is 10 years.)

 

[Unco]

...find where 1.08 to the nth power is closest to 2.00.

\(\displaystyle \text{1.08 to the nth power \, \Leftrightarrow \, 1.08^n\)
....\(\displaystyle \text{is equal to 2.00 \, \Leftrightarrow \, 1.08^n = 2.00\)

Solve by taking logs (any base will do, here I will use log base 10 on my calculator):

\(\displaystyle \log{(1.08^n)} = \log{(2)}\)

One of the log rules: \(\displaystyle \log_b{a^n} = n\log_b{a}\)

So we have

....\(\displaystyle n\log{(1.08)} = \log{(2)}\)

Hence

....\(\displaystyle n = \frac{\log{(2)}}{\log(1.08)} = 9.006\) (4sf)

So you were spot on for the first one.

You can do the second now.

 

[Gene]

Or you can use the rule of 72. The number of compoundings to double your money at x% = about 72/x% so 72/8 = 9