I needed to see if something to be improved twice, i.e. $100%$ increase, by increasing by $1%$ at a time, how long it will take. I figured that this is like compound interest where $1%$ applied each day, if so when will I double my money. I was able to deduct the formula to the following knowing the interest and the fact that it needs to double the initial amount.
$2 = 1.01^n$.
I need to now know n, how do I move forward. It's been ages since I done some maths. Thanks for your help.
1 Answer
As you point out, you need to solve the equation $(1.01)^n=2$. Almost always, there will not be an exact integer solution.
There are two possible approaches:
$1.$ Fool around with your calculator, raising $1.01$ to various powers. Pretty soon, you should be able to locate your $n$.
$2.$ Use logarithms. Any base will do, as long as you are consistent.
We have $(1.01)^n=2$ if and only if $n \log(1.01)=\log(2)$. So the exact real number $n$ such that $(1.01)^n=2$ is given by $n=\dfrac{\log(2)}{\log(1.01)}$.
If you want an integer answer, round to the nearest integer.
The same idea will take care of similar problems. For example, if in time $1$, a quantity increases by $2\%$, then the amount of time for the quantity to grow by $50\%$ is $\dfrac{\log(1.5)}{\log(1.02)}$.
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$\begingroup$ You are welcome. You had the idea basically right. The general fact about logarithms that I used is $\log(a^b)=b\log(a)$. $\endgroup$ Mar 11, 2013 at 7:16
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$\begingroup$ hi.. it does ring some bells now... Maths is truly amazing.. :) $\endgroup$– xelberMar 11, 2013 at 21:58
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$\begingroup$ Unfortunately, all too easily forgotten if one is away from it for a while. But recoverable. Logarithms were a great discovery, revolutionized astronomy. Computations that were once too painful to carry out became feasible. $\endgroup$ Mar 11, 2013 at 22:12