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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.

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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: – 1015 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
@rlgordonma: not quite. If $y = x^{1/40}$ you get $y^{40} + y^{37} = 15$. – Robert Israel Jan 25 '13 at 21:21
up vote 3 down vote accepted

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.

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@ncmathsadist Thanks for the edit. Some people call the bisection method/algorithm the dichotomy method: – 1015 Jan 25 '13 at 21:50

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.

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