I don't!
thanks very much for that table, it makes things a little clearer...
elouise
If you return, a couple of thoughts.
(1) I agree with how Halls read the question. To show the second month's sales as 1212 means a growth rate of 102% a month. Pretty implausible.
(2) Either way, the problem is easy to solve using a spread sheet like excel. Unfortunately, using excel will not teach you much math. (Of course, with spread sheets around, fewer people may need much math.)
(3) If you have a Windows machine ( and probably if you have an Apple), it will have a built-in calculator that can be set to scientific mode. You will then have either an e
x or a ln key (probably both). In my version of Windows, the calculator is under Accessories, and you set it to scientific mode by clicking on View.You have to toggle between e
x and ln. If you do not understand these directions, please ask for clarification.
(4) I am going to expand a bit on Hall's mathematical answer.
In math notation
\(\displaystyle a^0 = 1.\) This can also be shown as a^0 = 1.
\(\displaystyle If\ n\ is\ a\ positive\ whole\ number,\ a^n = a * a^{n-1}.\) This can also be shown as a^n = a * a^(n - 1).
Let's calculate April's sales the hard way.
\(\displaystyle 600 + 600 * 2\% = 600 + 600 * \dfrac{2}{100} = 600 * \left(1 + \dfrac{2}{100}\right) = 600 * \dfrac{100 + 2}{100} = 600 * 1.02 = 612.\)
With me so far?
Let's calculate May's sales the hard way
\(\displaystyle 612 + 612 * 2\% = 612 + 612 * \dfrac{2}{100} = 612 * \left(1 + \dfrac{2}{100}\right) = 612 * \dfrac{100 + 2}{100} = 612 * 1.02 = 624.24.\) Still with me?
But we could have done it more easily
\(\displaystyle 612 = 600 * 1.02 \implies 612 * 1.02 = (600 * 1.02) * 1.02 = 600 * 1.02^2.\)
Now we see that
\(\displaystyle March's\ sales = 600 = 600 * 1.02^0.\)
\(\displaystyle April's\ sales = 612 = 600 * 1.02 = 600 * 1.02^1.\)
\(\displaystyle May's\ sales = 624.24 = 600 * 1.02^2.\)
In other words, if we number March of this year as month 0
\(\displaystyle Sales\ for\ month\ n = 600 * 1.02^n.\) Make sense?
This gives a nice easy formula, and you can solve the problem fairly quickly by trial and error witha scientific calculator.
(5) Now if you know about logarithms, the problem becomes a simple equation where x is the unknown month when sales first equal or exceed 40,000
\(\displaystyle 600 * 1.02^x \approx 40,000 \implies 1.02^x \approx \dfrac{40000}{600} \implies ln\left(1.02^x\right) \approx ln\left(\dfrac{40000}{600}\right) \implies x * ln(1.02) \approx 4.1997 \implies 0.0198x \approx 4.1997 \implies\)
\(\displaystyle x \approx 212.1 \implies x = 213.\)
Let's check:
\(\displaystyle 600 * 1.02^{212} < 40000.\)
\(\displaystyle 600 * 1.02^{213} > 40000.\)
It is to solve problems like this that you learn logarithms.