+4 i wanted to know if there is an equation for adding a certain percentage onto the same thing over and over. an example would be 100 + 10% would be 110, however you can't just add 10 again, because, now 110+10% = 121, not 120, so i wanted know if there was an equation to solve problems like this in large amounts rather than having to use a calculator and tying each number in over and over (or mabey if someone had a website for something like this)
Yes, there is a shortcut to that:
100 x [1 + 10%]^n =100 x [1 + 0.10]^n=100 x 1.10^n. If n=20, then: 100 x 1.10^20=100 x 6.7275 =
672.75.
+245 Hey man. I just made a simple program that will execute what you asked for. Hope it helps. Click the link. Then click the button with arrows that point at each other. Then you should see a green block that you can put numbers into to do what you wanted.
http://snap.berkeley.edu/snapsource/snap.html#present:Username=jbocanegra&ProjectName=Percent%20
+118609 Hi JppDragon and thanks guys 
It is nice to meet you Dragon :)
What guest is saying is that you are not adding 10% each time, you are adding 10% of the previous total.
'of' automatically implies multiply.
You are COMPOUNDING the original amount - this is what compound interest is all about.
Say the initial amount is $90. This will be less confusing than using $100
A=$90
after 10% is added you will have
$90 + 10% of $90
=100% of $90+10% of $90
=110% of $90
which is the same as
= 90*(100%+10%)
=90*(1+0.1)
=90*1.1 = $99
If you want to increase it a SECOND time by 10% you will get
\($90 * 1.1 *1.1 = $90 * 1.1^2\)
f you want to increase it a THIRD time by 10% you will get
\($90 * 1.1 *1.1 *1.1 = $90 * 1.1^2*1.1=$90*1.1^3\)
Can you see the pattern?
Say you want $60 to be increased by 5% every your for 8 years. This is how it is done.
5% = 0.05
\(future\; value=$60*(1+0.05)^8\\ future\; value=$60*(1.05)^8\\ \)
60*1.05^8 = 88.64732662734375
So it will grow to $88.65 (to the nearest cent)
The fromula is
\(FV=PV(1+r)^n\)
where
FV=future value
PV=present value
r = rate expressed as a decimal
n = number of compounding intervals (time)
If you have any questions then just ask :)
