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

## Sunday, February 22, 2015

## Monday, February 16, 2015

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.

## Wednesday, February 11, 2015

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"

## Sunday, February 8, 2015

## Tuesday, February 3, 2015

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.

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