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#1
Hey. Can anyone help me solve the below for n?

0.012n2 + 0.108n = 1.12n - 1

I tried using logs but I don't seem to be getting anywhere

p.s. I know what the numerical answer is
Last edited by SapphirePhoenix1; 2 days ago
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2 days ago
#2
(Original post by SapphirePhoenix1)
Hey. Can anyone help me solve the below for n?

0.012n2 + 0.108n = 1.12n - 1

I tried using logs but I don't seem to be getting anywhere

p.s. I know what the numerical answer is
Not an idea as to how to solve - I'll have a think about that - but in your second line, you have used the idea that log(A + B) = log(A) + log(B). Is this correct?
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#3
(Original post by Pangol)
Not an idea as to how to solve - I'll have a think about that - but in your second line, you have used the idea that log(A + B) = log(A) + log(B). Is this correct?
Hey,

Yep, that's correct
0
2 days ago
#4
(Original post by SapphirePhoenix1)
Hey,

Yep, that's correct
No, it isn't!
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2 days ago
#5
(Original post by SapphirePhoenix1)
Hey. Can anyone help me solve the below for n?

0.012n2 + 0.108n = 1.12n - 1

I tried using logs but I don't seem to be getting anywhere

p.s. I know what the numerical answer is
There is two trivial answers ... Where does the question come from?
I'd be surprised if there was a traditional derived solution.
Last edited by mqb2766; 2 days ago
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#6
(Original post by SapphirePhoenix1)
Hey. Can anyone help me solve the below for n?

0.012n2 + 0.108n = 1.12n - 1

I tried using logs but I don't seem to be getting anywhere

p.s. I know what the numerical answer is
Update:
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2 days ago
#7
I can't think of any analytical way to do this. As mqb2766 says, there is a trivial solution - well, two - but graphing the function shows that there are three solutions altogether. It's possible that an itterative method would converge to that one with a suitable starting point. Perhaps someone cleverer than me can find an elementry way to solve this, but I can't see it, and mixtures of polynomials and powers don't often yield to them.
Last edited by Pangol; 2 days ago
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2 days ago
#8
(Original post by SapphirePhoenix1)
Update:
You've got the same problem as soon as you take logs. log(A + B) is not the same as log(A) + log(B)
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#9
(Original post by Pangol)
I can't think of any analytical way to do this. As mqb2766 says, there is a trivial solution - well, two - but graphing the function shows that there are three solutions altogether. It's possible that an itterative method would converge to that one with a suitable starting point. Perhaps someone cleverer than me can find an elementry way to solve this, but I can't see it, and mixtured of polynomials and powers don't often yield to them.
Yeah, I tried graphing it on my calculator and it gave me the numerical answer I was given by my teacher.

Thanks anyway
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#10
(Original post by Pangol)
You've got the same problem as soon as you take logs. log(A + B) is not the same as log(A) + log(B)
Yeah I know. I just changed log n2 in line 2 to 2logn
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2 days ago
#11
(Original post by SapphirePhoenix1)
Yeah I know. I just changed log n2 in line 2 to 2logn
Just to be sure - you do realise it's a bigger problem than this? You cannot go from 0.012n2 + 0.108n = 1.12n - 1 to log(0.012n2) + log(0.108n) = log(1.12n) - log(1).

Anyway, an iterative method does seem to work. If you rearrange to make the n which is tied up in the 1.12n term the subject of the equation and use a starting point larger than 1, it converges to the other solution.
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2 days ago
#12
As Pangol said, there isn't generally a solution to problems which mix exponential and polynomials. In this case, for small n, you have a couple of easy solutions by simply coefficient spotting,.. Then a larger one where the exponential catches back up with the quadratic as the base is just slightly >1.
Last edited by mqb2766; 2 days ago
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