Totally not homework I swear...

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.

Picture only slightly unrelated.

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montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U18_L4_T1_text_final.html
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>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.

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.

30640(1.015n)= 671289 without compounding
30640(1.015)^n = 671289 with compounding

Do you want to solve this using a computer or by hand?

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.

with my dick

I was gonna say: big number - little number = 0.015x, which is the same as your first thing.

Based on your second explanation you want to use the second one because it's compounding.

oh shit wait I think that's wrong depending on what op wants

>30640(1.015)^n = 671289 with compounding
k, thanks. I'll give that a shot.

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.

I realized that the minute I started looking into it.

I feel like my answer is somewhere on this page, montereyinstitute.org/courses/DevelopmentalMath/COURSE_TEXT2_RESOURCE/U18_L4_T1_text_final.html but I'm just not seeing it atm. Gonna keep digging around though.

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.

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

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...

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

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.

If you don't want to use a calculator, see log tables (base 10) at sosmath.com/tables/logtable/logtable.html

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

Hmm, seems awfully close, off by 1?
What was done wrong?

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.

I started my counting from 30640.

I'll give you my brute force list cutting off decimal points, if that helps.

30,640
31,099
31,566
32,039
32,520
33,007
33,503
34,005
34,515
35,033
35,558
36,092
36,633
37,182
37,740
38,306
38,880
39,464
40,056
40,656

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.

I get what the question is, I'm just wondering what was wrong with my algebra?

Sorry. Misunderstood.

I have no idea, this very quickly exceeded my level of math skill.

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.