Suppose you want to know how long it will take for an investment to double if it is compounded quarterly at 7%.

Use the formula A + P(1+ r/n)^(nt) where A is the amount, P is the initial investment, r is the interest rate, n is the number of times compounded annually and t is the time in years.

2A = A(1 + .07/4)^(4t)

2 = (1.0175)^4t

log 2 + log (1.0175)^(4t)

log 2 = 4t log(1.0175)

log2/log(1.0175) = 4t

t = 10 years

Another way to represent an exponential equation is by using logarithms. 

Suppose you have 5^x = 33 and you want to solve for x. That is difficult as it is set up. It can be made easier by using logarithms.

We rewrite the exponential equation above in logarithmic form as follow:

log (base b) y = x is the same ax b^x = y.

In our example, the logarthmic equivalent is log (base 5) 33 = x.

Using the change of base formula you get log 33/log 5 = x.  This can be solve on a calculator to get 2.1725. We can check this by substituting for x in the equation to get 5^(2.1725) which is approximately 30.

What is the difference between "highest 89 percent" and the "89th percentile" ?

There is a huge difference and one that is often confused.

The "highest 89 percent" is from the 11th percent on upward. In other words, if there are 100 people taking a test and you tare in the highest 89 percent, you have scored better than only 11 people.

The "89th percentile" would be that only 11 people scored higher. So it's the exact opposite.

"89th percentile" = "highest 11 percent"

"highest 89 percent" = "11th percentile"

Remember the key to consistency with math is to work problems daily. Master the technique of solving problems, know the steps to the correct solution. Do not just simply memorize steps, though, because it's important to understand what is going on in the problem.

Suppose we know that f"(x) = 5x^4 + 2x^3 + x + 3 and f'(0) = 4 and f(1) = 9.883333.  What is f(x)?

We have to integrate f''(x) first to get f'(x).

f'(x) = x^5 + (1/4)x^4 + (1/2)x^2 + 3x + c, since f'(0) = 4, we get

4 = 0^5 + (1/4)(0)^4 +(1/2)(0)^2 + 3(0) = c, therefore c = 4

f'(x) = x^5 + (1/4)x^4 + (1/2)x^2 + 3x + 4.  We get f(x) by integrating f'(x).

Therefore f(x) = (1/6)x^6 + (1/20)x^5 + (1/6)x^3 + (3/2)x^2 + 3x + c. Since we know f(1) = 9.883333, we get

9.883333 = 1/6 + 1/20 + 1/6 + 3/2 + 3  + c.   Therefore, c =5 and f(x) is

f(x) = (1/6)x^6 + (1/20)x^5 + (1/6)x^3 + (3/2)x^2 + 3x + 5.