maths help

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

(edited 3 years ago)

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Reply 1

Pangol

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?

Reply 2

SapphirePhoenix1

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 :smile:

Reply 3

RogerOxon

Original post by SapphirePhoenix1

Hey,

Yep, that's correct :smile:

No, it isn't!

Reply 4

mqb2766

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.

(edited 3 years ago)

Reply 5

SapphirePhoenix1

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:

Reply 6

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 mixtures of polynomials and powers don't often yield to them.

(edited 3 years ago)

Reply 7

Pangol

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)

Reply 8

SapphirePhoenix1

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 :smile:

Reply 9

SapphirePhoenix1

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

Reply 10

Pangol

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.

Reply 11

mqb2766

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.