I've gotten into logarithms in school, and it seems interesting to me. I enjoy working on it.

However, the curriculum I use (homeschool curriculum) isn't clear on a couple points, which I would like to ask about now.

First: My curriculum gave me a list of common logarithms and antilogarithms. The logarithms on the table range from 1.0 to 9.9 in 0.1 increments, and the corresponding antilogarithms range from .0000 to .9996.

When the antilogarithm required for an equation is not on the list, I was instructed to round to the closest one (for example, if I'm looking for antilog .4777, log 3.00 is .4771 and log 3.01 is .4786--so I would round to 3.00).

Well, when I saw the grade on my homework was an F, I couldn't believe it. I thought I did everything correctly. But when I looked at what the correct answers were, I realized what had happened.

Here's what one of the problems were, and the steps I took to get to the answer. At the end I put their answer to show that it is similar, but not exactly the same (i.e. they rounded their antilog differently than they told me to, I think).

PROBLEM

At his son's birth, a man invested $2,000 in savings at 6% for his son's college education. In 19 years, how much, approximately, will be available?

The formula is A = P(1 + i)^n

A = accumulated money

P = principal

i = interest

n = number of years

Substituting the values:

A = 2000(1.06)^19

Converting to logarithm (log10 is assumed):

log A = log 2000 + 19 log 1.06

Finding the values on the provided table and substituting them in:

log A = 3.3010 + .4807

log A = 3.7817

Convert to antilog to isolate the variable and take out the 3:

A = (antilog .7817)(10^3)

According to their table, log 6.05 is .7818 and log 6.04 is .7810. 6.05 is closer, so I used that value and multiplied by 10^3 to find A:

A = 6,050

The accumulated amount of money over 19 years will be $6,050

My answer was given an F. I looked at their answer: $6,049. Obviously they had used a different value for antilog .7817.

I figured out that in order to get $6,049 as the answer, antilog .7817 would have to be 6.049. But how would one figure out to use this logarithm?

Second question, which is really just the first one taken a bit further...

PROBLEM

(Using the value of A from the previous problem)

Because of inflation, money is depreciating in real value. The formula for the real value P of A dollars n years from now at the rate of i per year is P = A(1 + i )^-n. If the rate of inflation is 8% a year, what is the current real value of the A = 2,000(1.06)19 dollars from the preceding problem above? After 19 years, did the father and son have more or less money, as measured by current real value, than they did at the son's birth?

Using that formula, I can break it down after a few steps to this:

log P = 3.1472

P = (antilog .1472)(10^3)

There is no exact match in the table. 1.40 is .1461 and 1.41 is .1492. 1.40 is closer, so using that, I get the answer:

P = 1400

$1,400 is the real value of the saved money

Their answer is $1,403. In order to get to this answer, antilog .1472 would have to be 1.403...how could I possibly guess that?

Thanks for taking the time to read this huge problem, I hope I made it clear what the problem is and I hope you can help me! Please and thank you!