A little maths help please

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Could some kind soul please tell me how to solve this for x. It's been a long time since I did a-level maths, and can't recall if or how I ever did this kind of thing!

254 = 255x - x^12
 
calnen said:
That confused me as well :confused: If it's 'x to the power of 12' then it becomes rather more complicated.

yer i didnt misread it but just didnt look long enough, and im still in as level so i went by what i knew which is what i sed lol :p
 
doh. it is not a quadratic.

Basically I'm starting a fictional .com business earning £1000 in the first month with a total of £255,000 earnt by the end of month 12. How much would I have to make each month asuming the growth in linear?

I got this far

a + ar + ar^2 + ar^3.... ar^11= Sum

a = 1
Sum = 255
r = ?
n = 12

image057.gif
 
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If you think of it as a straight line graph would that help?

(I mean that if you plot (1,1000) and then (12,255000)).

So the gradient is:

Change in Y / Change in X

=

(255000 - 1000) / (12 - 1)

=

254000 / 11

=

23090.909090909090909090909090909

Therefore the equation of the linear growth would be:

y = 23090.909090909090909090909090909x + c

Where "c" is the Y intercept.

Therefore:

x = 1, y = 1000

1 * 23090.909090909090909090909090909 + c = 1000

c = -22090.909090909090909090909090909

So the equation is:

y = 23090.909090909090909090909090909x -22090.909090909090909090909090909

Then plug in the numbers: 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 for "x" to get the monthly figures.

Would that work? (I am not thinking terribly deeply here).

Angus Higgins
 
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First thing that strikes me is that x=1 is an obvious solution.

Second thing that strikes me is that if you differentiate it you will find that there's a unique turning point, so there can't be more than 2 solutions.

To find the second solution you can either do it numerically (it's a shade below 1.5) or you can use polynomial long division.
 
Angus-Higgins said:
If you think of it as a straight line graph would that help?

(I mean that if you plot (1,1000) and then (12,255000)).

Would that work? (I am not thinking terribly deeply here).

Angus Higgins

I'm not sure I follow. are you talking about graphing cumulative income? Because by the 12th month I want to have made 255k all together.
 
Topgun said:
doh. it is not a quadratic.

Basically I'm starting a fictional .com business earning £1000 in the first month with a total of £255,000 earnt by the end of month 12. How much would I have to make each month asuming the growth in linear?

I got this far

a + ar + ar^2 + ar^3.... ar^11= Sum

a = 1
Sum = 255
r = ?
n = 12

image057.gif
What you've posted there isn't linear growth, it's geometric (or exponential) growth. An example of linear growth would be 1000 in month 1, 2000 in month 2, 3000 in month 3 etc.

So you're looking for the number x such that

1000 + (1000 + X) + (1000 + 2X) + ... + (1000 + 11X) = 255,000

Which simplifies to

12,000 + 66X = 255,000
 
254 = 255x - x^12

x(255-x^11)= 254

therefore x= 254 and
255-x^11=254 ==x^11=1 therefore we can say x=1

x=1 and x =254 unless i am flawed in my thinking, it's been a few years since i've had to deal with these.
 
SpeedFreak said:
254 = 255x - x^12

x(255-x^11)= 254

therefore x= 254 and
255-x^11=254 ==x^11=1 therefore we can say x=1

x=1 and x =254 unless i am flawed in my thinking, it's been a few years since i've had to deal with these.
I think the flaw comes in the line where you start with "Assume x=254..." and end with "...and therefore x=1". ;)
 
Topgun said:
I'm not sure I follow. are you talking about graphing cumulative income? Because by the 12th month I want to have made 255k all together.

Ahhh, I think mine is wrong due to that; I didn't think it was cumulative. :p

SpeedFreak said:
254 = 255x - x^12

x(255-x^11)= 254

therefore x= 254 and
255-x^11=254 ==x^11=1 therefore we can say x=1

x=1 and x =254 unless i am flawed in my thinking, it's been a few years since i've had to deal with these.

Wouldn't it have 12 solutions being a 12th order equation?

Angus Higgins
 
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