How to solve this maths problem

[Bob the Planner]

I'm getting old and cannot remember how to solve this equation. Can some one give me the way to do it

300 = 10*Y^7 + 20*Y^4

I think I have to use logs to solve it due to the exponential but I cannot remember how (Too many years since my maths exams)

Ta

Bob

[Don]

y = 1.52505

I think.

Solved with Excel, rather than any mathematical knowledge you understand...

>> Edited by Don on Thursday 8th December 19:35

[Bob the Planner]

Don,

How did you get there ? Its the method I need not the solution to that particular problem

Bob

[aranell]

I don't think this is an easy problem to solve.
Being a seventh order polynomial, you can firstly try the rational roots (which we already know won't work, but it doesn't hurt to work through it)

Firstly, reduce the equation by dividing by 10: y^7 + 2y^4 - 30

constant = 30, meaning factors of 1,2,3,5,6,10,15,30
Leading coefficient = 1, meaning a factor of 1

So, rational roots will be ( + /-) 1/1, 2/1, 3/1, 5/1, 6/1, 10/1, 15/1, 30/1 there's a plus sign in the brackets before the numbers which doesn't show...you need to check the positive and negative of each number in the equation
Reduce fractions and try each root in the equation.

From this, we can see that there are no rational roots

From here, I think you need to use the approximation method instead. Firstly, graph the equation (don't ask me on this one...I hate graphing equations). Here, you will see the approximate roots. From here, you will have to "guess" the nearest two numbers, and then the nearest two tenths etc etc

I don't know of a 'formula' for seventh order polynomials, however I'm sure there are people on here who have much better maths skills than me and will probably set me straight

HTH,

- A

[2 Smokin Barrels]

At the risk of looking stupid a bit late to worry about that I know

* is multiply

= is equals

what does ^ mean please?

just curious

[scorp]

2 Smokin Barrels said:

At the risk of looking stupid a bit late to worry about that I know * is multiply = is equals what does ^ mean please? just curious

Raise to the nth power, i think.

[2 Smokin Barrels]

scorp said:

2 Smokin Barrels said: At the risk of looking stupid a bit late to worry about that I know * is multiply = is equals what does ^ mean please? just curious Raise to the nth power, i think.

That's what I was guessing.

[aranell]

Don said:

y = 1.52505 I think. Solved with Excel, rather than any mathematical knowledge you understand...

Incidentally, I suspect (but can't be certain) that there is more than one value for x which is why I suggest you graph the equation first to determine how many values there will be. The graph will also give you an indication of where these values lie.

- A

[gingerpaul]

Just spent some time trying to solve this and couldn't on paper.

A can tell you that there is only one solution and that this solution is roughly y=1.5250099355 although this has been kindly served up by my graphical calculator.

[chaparral]

There are seven roots. Six of them are complex conjugate pairs. The other is the one you guys have been getting...

[fido]

Bob the Planner said:

Don, How did you get there ? Its the method I need not the solution to that particular problem Bob

rearrange it into a nicer formula:
f(y) = y^7 + 2 y^4 - 30

you'll need the derivative:
df|dy = 7 y^6 + 8 y^3

it's a nice polynomial so with a bit of Newton recursion:
y_i+1 = y_i - f(y) / df|dy

start of with a guess:
y_i = 1

and after a few iterations (do it on Excel) you should get the above answer.

nope, it's still not cured my insomnia

[littlegreenfairy]

I just used maple to work it out and it gave me gibberish. I'm going to have a go using an ancient method called ye olde pencil and papyrus....

[xiphias]

x:=y^7 + 2*y^4 - 30;
7 4
x := y + 2 y - 30

> solve(x, y);

RootOf(%1, index = 1), RootOf(%1, index = 2), RootOf(%1, index = 3),

RootOf(%1, index = 4), RootOf(%1, index = 5),

RootOf(%1, index = 6), RootOf(%1, index = 7)

7 4
%1 := _Z + 2 _Z - 30

> evalf({%});

{-0.2710034536 - 1.534486795 I, 1.036448954 + 1.382928406 I,

-1.527950468 - 0.6148080103 I, 1.525009936,

1.036448954 - 1.382928406 I, -1.527950468 + 0.6148080103 I,

-0.2710034536 + 1.534486795 I}

> convert({%},fraction);

-16793 65541 40862 81283 -60215 16732 26860
{{------ - ----- I, ----- + ----- I, ------ - ----- I, -----,
61966 42712 39425 58776 39409 27215 17613

40862 81283 -60215 16732 -16793 65541
----- - ----- I, ------ + ----- I, ------ + ----- I}}
39425 58776 39409 27215 61966 42712

>

Happy?

Edit: Holy schmoke batman! Any way to make it monosize text?

>> Edited by xiphias on Friday 9th December 01:06

[Marshy]

All greek to me... I wish I did understand this stuff, but I lost it in the middle of my A-levels...

[SaTTaN]

I'm sure the answer should be 42?

It is the meaning of life, the universe and everything you know..

[Don]

Bob the Planner said:

Don, How did you get there ? Its the method I need not the solution to that particular problem Bob

Sadly, Bob, I simply used a computer. I just had a cell in Excel represent Y then created the formula. I didn't even bother to write a loop to iterate to the correct input value to achieve the output value of 300 - I just popped in 1 and 2 discovered the answer was between them and then fiddled about.

No mathematical knowledge required.

I am convinced, however, as you are that there is some direct method of solving this equation. After all - it only has ONE unknown: Y - but I can't remember how you deal with powers in an efficient manner...

[mindgam3]

300= 10*y^7 + 20*y^4

ln300 = ln10(y^7) + ln20(y^4)

ln300= 7ln10y + 4ln20y

ln 300 = 28ln200y

ln300/28 = ln 200y

200y = 300*exp(1/28)

y= 300/200*exp(1/28)

which gives you about 1.5 so its somewhere in the region of everyone else

Probably heaps of mistakes, i've only just got up, but you get the gernal jist

>> Edited by mindgam3 on Friday 9th December 08:09

[speedy_thrills]

Don said:

I am convinced, however, as you are that there is some direct method of solving this equation.

Newton-Raphson?

[beano500]

speedy_thrills said:

Newton-Raphson?

No! I haven't touched a drop.....

[Bob the Planner]

mindgam3 said:

300= 10*y^7 + 20*y^4 ln300 = ln10(y^7) + ln20(y^4) ln300= 7ln10y + 4ln20y ln 300 = 28ln200y ln300/28 = ln 200y 200y = 300*exp(1/28) y= 300/200*exp(1/28) which gives you about 1.5 so its somewhere in the region of everyone else Probably heaps of mistakes, i've only just got up, but you get the gernal jist >> Edited by mindgam3 on Friday 9th December 08:09

Thanks Ian - I knew I had to take logs but couldn't work out/remember how to do it.

Thanks to everyone else for their input - I think we all have about the same value.

Bob