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How can I solve the following equation? I really can't figure out how to solve it:

$x^{1/2}-x^{1/3} = 0$

Thank you.

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Please, try to make the title of your question more informative. E.g., Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. –  Julian Kuelshammer Dec 13 '12 at 15:34
If you have another question, please ask a new question, instead of editing an old one. –  George V. Williams Jun 13 '13 at 16:37

10 Answers 10

up vote 75 down vote accepted

$$\begin{eqnarray*}x^{1/2}-x^{1/3} &=& 0 \\ \\ \iff x^{3/6} - x^{2/6} &=& 0 \\ \\ \iff x^{2/6}(x^{1/6} - 1)&=& 0 \\ \\ \iff [x^{1/6} = 1\text{ or}\;x^{2/6} &=& 0] \\ \\ \iff [x = 1\text{ or}\;x &=& 0]\end{eqnarray*}$$

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Another thumbs-up from me! +1 –  Amzoti May 17 '13 at 1:14

Putting $x=y^6, y^3=y^2,y^2(y-1)=0\implies y=0$ or $1\implies x=0$ or $1$

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Observe that $6$ is chosen as lcm$(2,3)=6$ –  lab bhattacharjee Dec 13 '12 at 15:16

You might have heard this a thousand times, but the equation really does speak here!

What it says is that when you raise a number ($x$) to two different powers ($\frac 12$ and $\frac 13$) they turn out to be equal (their difference is $0$).

The only two numbers that come to mind are $0$ and $1$. $\Big[ \forall n \in \mathbb R, \;\; 0^n=0$ and $1^n=1 \Big]$

So the solution should be $x=0\ \text{ or } \ 1$

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+1 for the very nice first two paragraphs… but I almost took it back for the disclaimer! If an intuitive approach isn’t mathematically sound, the response shouldn’t be “ah well, accept that it’s not so good, and use an alternative solution”. The response should be “OK, so which bit isn’t mathematically sound, and how can we make it so?” With a little work in the third paragraph, your answer can become a solution that’s both mathematically sound and more intuitively clear than some of the other answers here. –  Peter LeFanu Lumsdaine Dec 14 '12 at 3:35
I think that this can be trivially made "mathematically sound" and it's a nice way to show how carefully reading and understanding what the problem asks one to show is important before one starts writing. –  Nik Bougalis Dec 14 '12 at 4:50

(Henceforth, $x$ is a variable denoting a non-negative real number)

$$x^{1/2} - x^{1/3} = 0$$ if and only if $$x^{1/2} = x^{1/3} $$ if and only if $$\frac{x^{1/2}}{x^{1/3}} = 1 \qquad \text{or} \qquad x^{1/3} = 0 $$ if and only if $$x^{\frac{1}{2} - \frac{1}{3}} = 1 \qquad \text{or} \qquad x = 0 $$ if and only if $$x^{1/6} = 1 \qquad \text{or} \qquad x = 0 $$ if and only if $$x = 1 \qquad \text{or} \qquad x = 0 $$

$$x^{1/2} - x^{1/3} = 0$$ if and only if $$x^{1/2} = x^{1/3} $$ if and only if (because both sides are positive) $$x = x^{2/3} $$ if and only if $$x^3 = x^2 $$ if and only if $$x = 1 \qquad \text{or} \qquad x^2 = 0 $$ if and only if $$x = 1 \qquad \text{or} \qquad x = 0 $$

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there are way too many " and only i"s. +1 though –  John Joy May 30 '14 at 16:33

$x^{1/2}-x^{1/3} = 0$






$x^2 = 0$ or $x-1=0$

$x = 0$ or $x = 1$

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Similar to the first answer.

$$x^{\frac{1}{2}}-x^{\frac{1}{3}}=0$$ Rewrite $x^{\frac{1}{2}}$ as something clever: $$x^{3/6}-x^{\frac{1}{3}}=0$$ (In case that's illegible, that's x^(3/6))

Now factor out $x^{\frac{1}{3}}$: $$x^{\frac{1}{3}}(x^{\frac{2}{6}}-1)=0$$

Now divide both sides by $x^{\frac{1}{3}}$. Note that this assumes $x \neq 0$; we will have to go back and check for that case later.

$$x^{\frac{2}{6}}-1=0$$ $$x^{\frac{2}{6}}=1$$ $$x^{\frac{1}{3}}=1$$ $$(x^{\frac{1}{3}})^3=1^3$$ $$x=1$$

Now let's go back and check if it works for $x=0$:


Yep, that equation holds!

So either $x=1$ or $x=0$.

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At first I saw $x^{\frac{3}{0}}$ and I almost downvoted! I proposed an edit to change the $\frac{3}{6}$ to to $3/6$. –  dimensio1n0 Jun 11 '13 at 4:33

More heuristic (in my opinion) solution: $$x^{1/2}=x^{1/3}$$ Immediately, realise that taking the logarithm base $x$ of both sides leads to $1/2=1/3,2=3,0=1,1=0$. So, $x$ must be such a number that you cannot take logarithm based that number? What are such numbers? $1$ and $0$. Plug both into the equation and notice that both work out well.

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$0$ is an apparent solution. There can be no negative solutions, given the $1/2$ power. So then assuming $x>0$, with logarithms:

$$\begin{align} x^{1/2}&=x^{1/3}\\ \implies \frac12\ln(x)&=\frac13\ln(x) \end{align}$$

So $\ln(x)=0$, making $x=1$ the only other possible solution.

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$x^{1/2}-x^{1/3} = 0$

$x^{1/3}(x^{1/2-1/3} - 1)= 0$

$x^{1/3}(x^{3/6-2/6} - 1)= 0$

$x^{1/3}(x^{1/6} - 1)= 0$

$x^{1/3}= 0$ and $x^{1/6} - 1= 0$

$x=0$ and $x^{1/6} = 1$

$x=0$ and $x = 1$

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The equation cannot have a negative solution since $x^{1/2}$ is defined in $R$ for $x>=0$ only.

$x=0$ is obviously a solution, since $0^n=0 \ \forall n \in \mathbb R$

For positive solutions, take logarithms on both sides of the equation $x^{1/2}=x^{1/3}$

It gives $log x=0 \implies x=1$

Hence the solutions to the equation are $x=0 , x=1$.

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