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/sci/ - Science & Math

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10215589 No.10215589 [Reply] [Original]

Is there an easy way to solve this without using logarithms?

I'm trying to help my cousin study for a test they won't be able to use calculators for, and they don't seem to have gotten into logarithms yet, but I can't remember any good way to solve this!

>Please help this brainlet that has been out of school for way too long!

>> No.10215597

>>10215589
Plug in 24 for t.....
You may want to ask someone else to help your cousin

>> No.10215599

>>10215597
>I'm trying to help my cousin study for a test they won't be able to use calculators for
>for a test they won't be able to use calculators for
>won't be able to use calculators
Did you read OP's post?

>> No.10215600

>>10215597
Yeah no shit, but how do you calculate something like 1,04^24 without a calculator??

>> No.10215605

>>10215599
>>10215600
They're not going to ask a question like that without allowing a calculator.

>> No.10215612
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10215612

>>10215589
[math] N(24)=6000\cdot 1.4^{25} [/math] there's nothing to solve for. If you mean evaluate, then common sense tells you that no teacher is going to ask for that without a calculator. Maybe you could use the binomial theorem if you want so many decimal points

>> No.10215615

>>10215605
She told me that calculators weren't allowed, and that this was a problem from a practice sheet she got for the test, which is why I'm asking, thinking I forgot some sort of trick or something

>>10215612
jfc dude

>> No.10215621

>>10215615
No tricks, unless they give the value of 1.04^24 or 1.04^23 or something in an earlier part of the question or something else like that

>> No.10215632

>>10215621
Nope. The first part of the problem asks "What does 1,04 and t mean in this equation?"
and then to find the value of y when t = 24.

It's a basic course of mathematics too, but the jump in difficulty, especially before handling things like logs, seems extreme.

>> No.10215647

>>10215632
Well logarithms wouldn't help here either. If you take the log of both sides you just get log(N(24)) = log(6000) + 24 log(1.04) which is hardly an improvement

>> No.10215662

taylor series/linear approximation

>> No.10216098

>>10215662
this

>> No.10216142

>>10215662
how

>> No.10216144

>>10216142
He's just memeing, if your cousin hasn't gotten to logarithms yet he's probably not going to know any calculus either

>> No.10216937

you could calculate it manually, it's not hard

>> No.10216969

>>10215589
Easiest way to calculate it is to do repeated squaring, 1.4^(2^4+2^3) but even that is unreasonable to do by hand, you just do manual multiplication for a while and that's it.

>> No.10216981

Alright so 1.04 = 1+ 0.4. Then note that (1+a)^2 ~ 1 + 2a for small a. (0.4 is small).
Then letting c = 1.04. c^2 = 1.08, c^4 = 1.16, c^8 = 1.32, c^16 = 1.64, then c^24 = c^(16 + 8) = (1+0.64)(1+0.32) ~ 1+0.32 + 0.64 = 1.96 ~ 2, then
6000 * 1.04^24 ~ 6000 * 2 = 12 000

>> No.10216983

>>10216981
alright now i went back and checked it with wolfram alpha and turns out it's not really close so nevermind my post.

>> No.10216987

>>10216969
Just work out what 2^48 is by hand, repeated squaring.
Then it's 1/(2^48) + 1 multiplied by 6000.

>> No.10216990

>>10216987
no it's not lmao go back to check your calculations

>> No.10216995

>>10216987
No no it's it's not, you might have to use linear algebra. Say you took a transition matrix to represent the squaring operation and then diagonalized it and then took the power of 24. Something like that.

>> No.10217001

>>10215589
unironically guess and check

>> No.10217006

note that 1.04 * 1.04 = 1.0816 and 1.0816 * 1.0816 = 1.16985856 ~1.17
let's do 6 rows of pascal's triangle
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1

then (1.04)^24 = (1.17)^6 = 0.17^6 + 6 * 0.17^5 + 15 * 0.17^4 + 20*0.17^3 + 15*0.17^2 + 6*0.17 + 1 ~ 15*0.17^2 + 6*0.17 + 1 = 2.4535 ~ 2.5

then (1.04)^24 * 6000 ~ 2.5 * 6000 = 15000

>> No.10217012

>>10217006
alternatively, you could do 12 rows of pascals triangle and only square once to get a more accurate result

>> No.10217032

>>10217006
No calculators allowed

>> No.10217045

>>10217032
wasn't using my calculator!!!!

>> No.10217051

>>10215589
test

>> No.10217202
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10217202

niggers just use differentials. 100% mad at you /sci/.

1.04

x = 1 (since you know that 1^24 = 1)
delta x = 0.04. since 0.04 is small,
delta x is approximately equal to dx.
So dx = 0.04.

dy/dx = d/dx x^24
dy/dx = 24x^23
dy = 24x^23 dx

for x = 1 and dx = 0.04 we have:

dy = 24*0.04 = 0.96.

Now the last part.

1.04^24 = y + dy
1.04^24 = x^24 + dy
1.04^24 = 1 + 0.96
1.04^24 = 1.96

1.04^24 is approximately equal to 1.96.

and 6000*1.96 = 11760

All able to be done without a calculator.
You should not care for an exactly equal answer without a calculator. If your teacher says you're wrong, call her a faggot. If your teacher says this is too advanced (Calculus 2) for your class and not what you should've done, call her a nigger.

Also, in the test calculators aren't able to be used, but maybe out of it (class, homework) you are expected to use it. Just that the test questions don't require a calculator.

>> No.10217224

>>10217202
You're answer for 1.04^24 is about three orders of magnitude off lmao

>> No.10217247

>>10215589
We can use the binomial formula
[eqn] (1+x)^{n} = \sum_{k=0}^n \binom{n}{k} x^k [/eqn]
since we can't be assed to calculate [math] \binom{n}{k} [/math] and [math] x^k [/math] for too big of a [math] k [/math], we'll do the approximation
[eqn] (1+0.04)^{24} \approx \sum_{k=0}^3 \binom{24}{k} x^k = 1+24\cdot 0.04+\frac{24\cdot 23}{2}0.04^2 + \frac{24\cdot 23\cdot 22}{6}0.04^3 [/eqn]
Which is:

[math] 1+24\cdot \frac{4}{100}+276 \frac{16}{10000} +2024 \frac{64}{1000000} [/math]
[math] = 1+ \frac{96}{100}+ \frac{4416}{10000} + \frac{129536}{1000000} [/math]
[math] = 1+ 0.96+ 0.4416+ 0.129536 = 2.531136 [/math]

The exact solution is [math] 2.563304 [/math] so the error is [math] 0.032168 [/math]
For comparison we have
[eqn] \left| 1.04^{24} - \sum_{k=0}^2 \binom{24}{k} x^k \right| = 0.161704 [/eqn]
and
[eqn] \left| 1.04^{24} - \sum_{k=0}^4 \binom{24}{k} x^k \right| = 0.004965 [/eqn]

>> No.10217255

>>10217224
He's probably a physicist

>> No.10217258

>>10217224
Guess that 0.04 wasn't small. You need to find a closer x like knowing already something like the result of 1.039^24 and then choosing x = 1.039 and dx = 0.001. Or a even closer result since the exponent is 24. But that isn't supplied in the question.

>> No.10217265

>>10217258
or you just use more terms like in >>10217247

>> No.10217270

>>10215632
You better be trolling if it truly is a super basic course then the answer is gonna be super basic and they dont want you to do anything else than put 24 in place of the t

>> No.10218084

>>10215600
1,04^24 = (1,04^8)^3 = (((1,04^2)^2)^2)^3
So five multiplications.

>> No.10218110

>>10215589
(1.04)^24 = (1 + 1/25)^24 = (1 + 1/25)^25 / 1.04 which is approx e * 0.96 so that answer is close to 6000 * 2.72 * 0.96 = 15667.

>> No.10218152

>>10218084
was just about to shit on you because the last multiplication would be a total pain in the ass without rounding, but I just checked it and even if you round to 2 decimal places after each multiplications you'll only have a relative error of 0.7% in the end.

This solution is great as well:
>>10218110
with a relative error of only 1.9%