Totally not homework I swear, but I have no real understanding of mathematics outside of basic Algebra and I want to try and figure something out.
I have two numbers, 671,289 and another 30,640. In my scenario, 30,640 increases by 1.015 each time.
without doing the brute force, (I realize I probably could have just brute forced it by now,) I would like to know of an equation I can use to figure out how many times would I have to let the smaller number 30,640 multiply by 1.015 before the total sum is equal to or greater than the 671,289. I'm pretty sure I can solve the problem on my own if I just have an equation to use.
>I would like to know of an equation I can use to figure out how many times would I have to let the smaller number 30,640 multiply by 1.015 before the total sum is equal to or greater than the 671,289. This sentence literally fried my brain, thanks for that, now I'm retarded.
Isaac Gray
Sorry, I'm really bad at explaining things.
I start off with 30,640. It get's multiplied by 1.015, getting me 31,099. 31,099 gets multiplied by 1.015 giving me 31,565.
Now if I stopped the string there and added the numbers I would get 93,395. How many times would I have to let the multiplication process continue before every result added together equaled 671,289.
I know theres an equation for this to figure it out on my own without just simply brute forcing it. I'd like to know the equation but I have no mathematics training or knowledge to look up what such an equation is.
Dylan Parker
30640(1.015n)= 671289 without compounding 30640(1.015)^n = 671289 with compounding
Jose Sanders
Do you want to solve this using a computer or by hand?
Jack Gray
If it's not terrible I'd like to do it by hand.
I kinda started this little trip and I don't really like giving up. math excites me, but I'm really bad at it.
Jayden Flores
with my dick
Joseph Mitchell
I was gonna say: big number - little number = 0.015x, which is the same as your first thing.
Logan Williams
Based on your second explanation you want to use the second one because it's compounding.
Jonathan Cox
oh shit wait I think that's wrong depending on what op wants
Easton Garcia
>30640(1.015)^n = 671289 with compounding k, thanks. I'll give that a shot.
Adam Anderson
This problem is terrible to do by hand.
Also, what you are looking for is not really a formula, but rather an algorithm. A formula in math is a description. An algorithm is a method to compute something.
Practically speaking, the most efficient algorithm for computing the answer to the stated question is brute force.
There are formulas for the answer, but the algorithm for computing the answer based on those formulas is much more involved. However, I would recommend learning about the formulas if you are interested in math. What you want to read into is the concept of exponentiation, the exponential function, and the logarithm.
Owen Reyes
I realized that the minute I started looking into it.
If you're interested, I'd be willing to chat with you over discord to give an explanation. You can add me at scout#6748.
Brandon Wright
Nigga what
Small number*multiplier^X = big number
Multiplier^X=big/small
X = log of big/small in base multiplier
Plug dat shit on the calc and fuck it i kms if u arent bait
Gabriel Hernandez
Your first OP post didn't make any sense, but your second explanation is what I'm going on.
This isn't solving it algorithmically, is this right Veeky Forums?
I got 18.0851...
Oliver Smith
I brute forced the equation and found I had to let it run 20 times.
19 times produced 667,838 and 20 produced 708,494
I'm sorry, it's not bait. I never learned log in High school, I essentially flunked out of geometry
Juan Rivera
Sorry, but that's wrong. I think you slipped a decimal.
30640 is multiplied by 1.015 again and again and again until it reaches 671289. This is a classic "compound interest" problem. Write it as 30640 * 1.015^N = 671289 where 1.015^N means 1.015 raised to the N-th power. Rearranging 1.015^N = 671289/30640 = 21.909
You raise a number to the Nth power by taking the logarithm of the number, multiplying it by N, and then taking the anti-logarithm.
Log of 1.015 = .00646604 Multiply by N and take the anti-log. Wait! We don't known N. We get around that by taking the log of 671289 on the other side of the equation. Log of 2.1909 is 0.3404441 Right digits, but we want the log of a larger number. That's why I referenced base 10 logs. Everytime the number is multiplied by 10, the logarithm increases by 1. We move the decimal 1 place and see the log of 2.19 (as close as we can get with these tables) is 1.3404441
Log(1.015) * N = log(21.909) N = log(31.909)/log(1.015) = 1.3404441/.00646604 = 201.73 The answer. Properly speaking, I should have interpolated in the tables, not just used the closest printed numbers. A calculator is easier. I get 207.33
Hudson Wright
Hmm, seems awfully close, off by 1? What was done wrong?
Christopher Lee
You've misunderstood OP. Perhaps read his better explanation: He wants to sum all the 30,640*(1.015)^n until that sum is equal to 671,289 and find what the last 'n' would have to be.
Hunter Hernandez
I started my counting from 30640.
I'll give you my brute force list cutting off decimal points, if that helps.
Add them all together and I got 708,494. Which brought me to my original question of how many times did I have to let it run till it was equal to or greater than 671,289.
Luis Nelson
I get what the question is, I'm just wondering what was wrong with my algebra?
Isaac Diaz
Sorry. Misunderstood.
Hudson Jackson
I have no idea, this very quickly exceeded my level of math skill.
Andrew Roberts
Use a geometric series!
The sum for a geometric series to n terms is given by Sn=(a*(1-r^n)/(1-r) where some term Un in the sequence is ar^(n-1)
So you start with some value a (30640) and each term on is the previous one times the common ratio r (1.015). n is the number of terms you have to sum.
So 30640(1-1.015^n)=671289(1-1.015)
Rearranging:
1.015^n=1-(671289/30640)(1-1.015) n*ln(1.015)=ln(1.3286336...) n=ln(1.3286336...)/ln(1.015) n=19.085... So you'd need to do the operation 19 times, summing 20 numbers in total, to get a number greater than it.