[katerie]

I can't seem to grasp the example given in this course, as I am doing it through correspondence. Can someone please help me understand how to solve this type of word problem?

Here's the questions:

A wise investor purchased a plot of land in 1947 for $84 000. In 1987 that same investor sold the land for $49 000 000. What annual rate of interest corresponds to an investment of $84 000 which grows to $49 000 000 in 40 years?

[ebaines]

You are trying to solve the following equation for "i," which is the interest rate that is assumed to be constant and compounded annually through the 40 years of this investment:

49,000,000 = 84,000*(1+i)^40

Rearrangeto get "i" by itself:

I = (49000000/84000)^(1/40) - 1

[katerie]

You are trying to solve the following equation for "i," which is the interest rate that is assumed to be constant and compounded annually through the 40 years of this investment:

49,000,000 = 84,000*(1+i)^40

Rearrangeto get "i" by itself:

i = (49000000/84000)^(1/40) - 1

Okay, I think I understand that, except I'm supposed to be solving using logarithms.

[ebaines]

If you start with an equation of the form:

A = B(1+i)^n

You can rearrange:

A/B = (1+i)^n

Take the log of both sides, and work from there:

ln(A/B) = n * ln(1+i)
1/n [ln(A) - ln(B)] = ln (1+i)

exp{1/n [ln(A) - ln(B)]} -1 = I

Here I used natural logs (base e), but you could just as easily have used log base 10 or any other base for that matter.