Solving an equation $x + x^{0.925} = 15$

Question

I have been trying to solve this equation for a couple of days now but am getting stuck.

$x + x^{0.925} = 15$, find $x$.

I went in the direction of taking log on both sides of the equation but that does not help and I cannot simplify further. Also, I started substituting values to find a range within which it might fall but am thinking there should be a better way to do that.

Any suggestions how to go about it. I just need a clue and I can work it out then.

One could try Newton's Method. Your tag says "Calculus"--If you're in CalcI, Newton's method could be the way to go. Because Wolfram Alpha only gives a numeric solution, I would say this is the route to take. — apnorton, Jan 25 '13 at 21:12
To begin with: http://www.wolframalpha.com/input/?i=x%2Bx%5E%280.925%29%3D15 — Julien, Jan 25 '13 at 21:13
According to Mathematica there isn't a closed form for the solution. What kind of answer are you looking for? — Antonio Vargas, Jan 25 '13 at 21:13
You can use the Intermediate Value Theorem to show that the real solution is between $8$ and $9$, say. — Robert Israel, Jan 25 '13 at 21:15
The exponent is a rational number ($37/40$), so maybe you wind up with an equation like $x^{40} + x^{37} = 15^{40}$. — Ron Gordon, Jan 25 '13 at 21:18
@rlgordonma: not quite. If $y = x^{1/40}$ you get $y^{40} + y^{37} = 15$. — Robert Israel, Jan 25 '13 at 21:21
@RobertIsrael: I see, then you have to raise your solution to the 40th power. Fun all around! — Ron Gordon, Jan 25 '13 at 21:26
@GauravSinha: good question! Newton's method, perhaps, although the high degree is cause for concern. — Ron Gordon, Jan 26 '13 at 15:14

score 3 · Accepted Answer · edited Jan 25 '13 at 21:35

3

Let's study the function $f(x)=x+x^{0.925}-15$, on its domain $(0,+\infty)$.

The derivative is $f'(x)=1+0.925x^{-0.075}$, which is positive for all $x>0$.

Since $\lim_{x\rightarrow 0^+} f(x)=-15$ and $\lim_{x\rightarrow +\infty} f(x)=+\infty$, it follows that $f$ is a bijection from $(0,+\infty)$ onto $(-15,+\infty)$.

In particular, there exists a unique $x>0$ such that $f(x)=0$.

Now $f(8)=-0.16...$ and $f(9)=1.63...$, so your zero is somewhere between $8$ and $9$.

If you want more precision, use the bisection algorithm.

edited Jan 25 '13 at 21:35

ncmathsadist

49,383

answered Jan 25 '13 at 21:25

Julien

44,791

@ncmathsadist Thanks for the edit. Some people call the bisection method/algorithm the dichotomy method: http://en.wikipedia.org/wiki/Bisection_method – Julien Jan 25 '13 at 21:50

score 0 · Answer 2 · answered Jan 25 '13 at 23:21

0

Another alternative. Since 0.925 is close to 1, x should be little higher than 15/2. Assume 8. Expand x^{0.925} around 8, i.e. $8^{0.925} + 0.925*8^{-0.075}(x-8)$. Now the equation becomes $x+6.84476+0.79143(x-8) \approx 15$. Solve x.

answered Jan 25 '13 at 23:21

karakfa

2,675

Solving an equation $x + x^{0.925} = 15$

2 Answers2