Hi, How would I solve to find x in 967.261271503x^{5.45} + 0.0000403225x = 0.00000077575?

Hi, How would I solve to find x in 967.261271503x^{5.45} + 0.0000403225x = 0.00000077575?
I could probably write a computer program to find the answer, or at least as close as need be. I think the program could also quickly determine, up front, if there is a solution. College math is behind me many, many years. I'm am very curious if and how this would be solved more "mathematically".
Oh you mean it's degree maths, and no quick way to find out?
No. Sorry to confuse you. I don't know how to solve it mathematically. It may be simple. It may be difficult.
You can use this scientific calculator to plug in different values for 'x' and see where it takes you: Online Scientific Calculator  EEWeb
I am unclear if the power 5.45 is just on the x or on the complete 967.261271503x
So does the distributive law of exponents apply here?
I assumed it was just on the 'x'. It has the "look" of a quadratic equation, except for the fractional power of 5.45 instead of 2.
Is it possible to first multiply both numbers so that the power is a whole, and then redivide the result after you've done the calculation?
Or does that not work with this kind of equation?
If you want to solve it numerically, just apply Newton's method and you should be able to find the answers to a reasonable degree of accuracy pretty quickly. High degree polynomials don't generally have nice solutions and noninteger powers are generally less nice.
Let there be a primitive algorithm to solve any polynomial equation. Then
967.261271503x^5.45 + 0.0000403225x = 0.00000077575
967.261271503x^5.45 = 0.00000077575  0.0000403225x
Then, raise both sides to the power of 20. The equation turns to something of the form:
A(x^109) = P(x)
P(x) is a polynomial of degree 20 and A is a constant (967.261271503^109, to be precise), multiplied by x^109. For some unknown reason it looks like a function A(x) on A(x^109). But it is actually a constant multiplied by x^109.
A(x^109)  P(x) = 0
This is a polynomial equation of degree 109. With a primitive algorithm to solve any polynomial equation, it can be solved. However with the primitive algorithms only of the axioms of the real numbers, it probably can't be solved. There is no general solution to polynomial equations of degree higher than 4, let alone degree 109, so unless this is a special case, it can't be solved using the facts we currently know about the real numbers.
Let there be a primitive algorithm to solve any polynomial equation. Then
967.261271503x^5.45 + 0.0000403225x = 0.00000077575
967.261271503x^5.45 = 0.00000077575  0.0000403225x
Then, raise both sides to the power of 20. The equation turns to something of the form:
A(x^109) = P(x)
P(x) is a polynomial of degree 20 and A is a constant (967.261271503^109, to be precise), multiplied by x^109. For some unknown reason it looks like a function A(x) on A(x^109). But it is actually a constant multiplied by x^109.
A(x^109)  P(x) = 0
This is a polynomial equation of degree 109. With a primitive algorithm to solve any polynomial equation, it can be solved. However with the primitive algorithms only of the axioms of the real numbers, it probably can't be solved. There is no general solution to polynomial equations of degree higher than 4, let alone degree 109, so unless this is a special case, it can't be solved using the facts we currently know about the real numbers.
Wow, thanks a lot guys, so many repliers.
the power 5.45 is indeed on the x.I am unclear if the power 5.45 is just on the x or on the complete 967.261271503x
I tried getting my ALevel maths teacher to solve this but he said he couldn't algebraicly, he told me to write a program to work it out numerically as well.If you want to solve it numerically, just apply Newton's method and you should be able to find the answers to a reasonable degree of accuracy pretty quickly. High degree polynomials don't generally have nice solutions and noninteger powers are generally less nice.
I believe so. This equation results from the use of solid of rotation to find the volume of a solid object. Then I halved the volume and pluged that back into the integral in order to find the new boundary which corresponds to half the volume. The equation above is precisely that. In fact I made a mistake, it should be = 0.00000077575/pi (which is just a number.) Anyway this means that the answer will be bellow 0.015 (because that's the upper boundary for the whole volume.) Hope that makes sense.Are you limiting yourself only to real values of x? The general collection of solutions is quite large...(although you are guaranteed at least one real solution here).
If you don't need to know how to solve it, you can just look it up:
http://www.wolframalpha.com/input/?i=solve+967.261271503*x^5.45+%2B+0.0000403225x+%3 D+0.00000077575
I would like to know how. That stuff anticorncob was writting looks interesting. Also how reliable is wolfram?
Lol, wolfram said standard computational time exceeded.
Wolfram is Mathematica.
It is industrial strength mathematical programming and pretty much the gold standard in the field.
It is a bit expensive for just hobby use at about $150.00 for the student edition or about $300.00 for the home edition.
Industrial and commercial versions cost a lot more.
Wolfram Mathematica: Technical Computing Software—Taking You from Idea to Solution
4^{3/2} = either 8 or 8. Which is, √(4^{3}) = (4^{3})^{1/2} should help you understand it; 4^{3} = 64, and 64^{1/2} = √64 = 8 or 8.
So raise the base to the power of the numerator of the fractional power, then take the denominatorth root of the result.
5.45 is 545/100 and you may reduce the fraction before taking the powers and roots. It's 109/20. Thus you take the 109th power of the base, and then take the 20th root of the result. That's what raising to the 5.45 power means.
Also its commutative; you can take the 20th root of the base and then the 109th power of the result, if that's more convenient. Same thing.
If you want to program it, use Newton's method. There are other ways to do the same thing faster, but Newton's method is simple. Just rewrite your equation in the form f(x) = 0, then plug it in to the formula given on the Wiki site. Just be aware that fractional powers often give imaginary results, so you may need a complex number class.
True, but that does not mean one can't get useful results from Newton's method. It does help if one has at least a rough idea of the value of the root(s) in question. Using that initial estimate as a seed frequently helps a great deal (although convergence is still not guaranteed).
In more difficult cases, I find that Laguerre's method does a remarkably good job. It seems (almost) always to converge, independent of the seed value, and converges relatively quickly.
Actually, Newton's Method is Chaotic.
Which turns out to be very cool; it's a fractal function that creates four basins of attraction depending on your choice of initial estimate. ETA: Yes, yes, I know about the number of complex roots. I'm talking about the simplest case and we all know it spans the class.
But it also makes it Turing and Kolmogorovundecidable.
Not that that's ultimately all that freaky, what we're really saying is we can't differentiate between 8 and 8 for certain in reverse physics of that type; which is merely affirming the 2LOT.
One does have to confess that raising to the 109th power is not trivial.
Actually, in this case, where you have 5 significant figures in the least accurate input, there's no point in calculating the root out beyond 5 figures, which should take less than 10 iterations. It'd probably only take 2 or 3 more to hit the 12 figures of the most accurate input. After that, any more digits you get are worthless. And in general, you'll fill out the significant figures of a floating point value pretty quickly.
As for it's chaotic nature, that's true in general, but not in every case.
I'm not saying there's aren't better methods, but you should try Newton's method first just because it's so simple. (And any numerical root finder will be nonterminating.)
Heh, absolutely, you're correct, and that's close enough for most work. I once upon a time bought a calculator that could numerically calculate a definite integral; it was the newest hottest thing.
I bought a graphing calculator six or eight years ago that blows it off the planet.
I never bothered with any of the computer solutions; my TI does it fine. If I did the stuff for a living I might drill deeper.
Thanks a lot guys. I'll get cracking.
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