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$$r\left(\frac 1p -1\right)- \left(\frac 1p\right)^\frac89 + \left(\frac 1p\right)^\frac19 =0$$

where $ r = 2^\frac89 - 2^\frac19$

How do I solve this without a computer?

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    $\begingroup$ What about $p=\frac{1}{2}$? If you choose $p$ so that $\frac{1}{p}=2$, then you'll find that the second and third terms match $r$, and, moreover, the factor on $r$ of $\left(\frac{1}{p}-1\right)$ becomes $1$. $\endgroup$ – Michael Burr Oct 30 '15 at 16:19
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Since $r=2^{8/9}-2^{1/9}$, it seems like it might be possible to choose $\frac{1}{p}=2$, i.e., $p=\frac{1}{2}$. This makes the expression become $r(2-1)-2^{8/9}+2^{1/9}=r-r=0$. Therefore $p=\frac{1}{2}$ is a root.

If $\frac{1}{p}=1$, then the first term vanishes and $1^{8/9}=1$ and $1^{1/9}=1$. Therefore $1$ is another real root.

As @user44394 states, observe that $\frac{1}{2}(2^{8/9}-2^{1/9})=2^{-1/9}-2^{-8/9}=(\frac{1}{2})^{8/9}-(\frac{1}{2})^{1/9}$ giving the third (and last by Descartes rule of signs) root of $p=2$.

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    $\begingroup$ Similarly to the case where $\frac1p = 2$, if $\frac1p = \frac12$, i.e., $p=\frac12$ then we find that $(2^\frac89 -2^\frac19)(-\frac12)=\frac12^\frac89-\frac12^\frac19$ and the LHS =RHS so is $p=2$ not also a solution? $\endgroup$ – Kai_M Oct 30 '15 at 16:39
  • $\begingroup$ @user44394 Nicely spotted. $\endgroup$ – Michael Burr Oct 30 '15 at 16:43
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First of all, there no need to express as 1/p so

$(2^{\frac 8 9} - 2^\frac 1 9)(m - 1) - m^{\frac 8 9} + m^{\frac 1 9} = 0$

$(2^{\frac 8 9} - 2^\frac 1 9)(m - 1) - (m^{\frac 8 9} - m^{\frac 1 9}) = 0$

Oh look! The $(m^{\frac 8 9} - m^{\frac 1 9})$ looks like $(2^{\frac 8 9} - 2^\frac 1 9)$! What if $m = 2$? Then $m - 1$ = 1 and we'd get an equality. So $p = \frac 1 2$ is a solution.

But that was us seeing something. Ho do we solve in general?

$(2^{\frac 8 9} - 2^\frac 1 9) = \frac{ (m^{\frac 8 9} - m^{\frac 1 9})}{m-1} $

$2^{\frac 1 9}(2^8 - 2) = \dfrac {m'^8 - m} {m'^9 - 1}$ for $m' = m^{\frac 1 9}$....

You know what? let's just say $p = \frac 1 2$ and call it a day.

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  • $\begingroup$ Also, include the easy $m=1$ answer. And +1 for making me chuckle. $\endgroup$ – Michael Burr Oct 30 '15 at 16:38

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