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The problem is as follows:

Given:

$$x^{-x}=(-8)^{-3^{-1}} \cdot (-8^{0})^{3^{4^{5}}}$$

Find the result of

$$(x^{-2}-x^{-1})$$

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&\frac{3}{4}\\ 2.&\frac{1}{4}\\ 3.&-\frac{1}{4}\\ 4.&\frac{1}{8}\\ 5.&\frac{3}{8}\\ \end{array}$

What I attempted to do was as follows:

From what I can understand this whole expression equals to $-1$ as the sign is taken out of the exponential and the $8$ powered to the rest becomes just $1$.

$(-8^{0})^{3^{4^{5}}}=-1$

Then if I'm not mistaken the thing is with the rest:

$(-8)^{-3^{-1}}=(-8)^{-\frac{1}{3}}=-\left(\frac{1}{2^3}\right)^{\frac{1}{3}}$

$-\left(\frac{1}{2^3}\right)^{\frac{1}{3}}=-\frac{1}{2}$

Then this means:

$\left(-\frac{1}{2}\right)\times (-1)=\frac{1}{2}$

Then this means:

$x^{-x}=2^{-1}$

But this doesn't seem right. I'm stuck there. Can someone help me here?. Maybe is my interpretation of the problem not accurate?. The official answer is the first option but I don't know how to get there.

Chris Steinbeck Bell
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    Numerically checked, none of the above is the solution. Typo? – Michael Hoppe Nov 23 '20 at 19:57
  • @MichaelHoppe See my comment following the mathSE answer given. Presumably, $x$ is a transcendental number. – user2661923 Nov 23 '20 at 20:11
  • @MichaelHoppe I have reviewed many times my steps and the computations made by hand and so far I couldn't spot any error. My best guess is that there could be a typo but since it has been mentioned that $x$ could be a transcendental number who knows. If there was some error I wonder what could it had been? maybe the intended power of the question was a power of two equal radicals or something? – Chris Steinbeck Bell Nov 24 '20 at 01:20
  • @user2661923 Because of this entry on wikipedia I have been introduced to transcendental numbers but again as I mentioned in my comment from above. It seems that it might had been a typo or who knows?. Have you spotted an error in my steps? – Chris Steinbeck Bell Nov 24 '20 at 01:21
  • No, I spotted no error. As near as I can figure, there are only two possibilities: (1) You misquoted the problem which is asking for which of the choices is closest to the value of $x$. (2) The problem composer was negligent, and that was his intent. – user2661923 Nov 24 '20 at 01:24
  • @user2661923 Well, I copied it down exactly as it was shown in my book. Its strange because most problems I could solve them right away but this one has gave me go in circles. No pun intended. Hence I am asking for assistance. – Chris Steinbeck Bell Nov 24 '20 at 01:31
  • @ChrisSteinbeckBell email your teacher (or somehow communicate with him) the link to this webpage, with an explanatory note, and see what he says. – user2661923 Nov 24 '20 at 01:37

2 Answers2

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If $$x^{-x}=8 \cdot (-8)^{-3^{-1}} \cdot (-8^{0})^{3^{4^{5}}}$$ or

$$x^{-x}=-(8)^{\left(\frac{3}{2}\right)^{-1}} \cdot (-8^{0})^{3^{4^{5}}},$$

then we have $$(x^{-2}-x^{-1})=\frac{3}{4}.$$

YNK
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x^x = 2 Equalize the options in the book (1-x)/(x^2) and solve the equation. And put the obtained answers in the equation x^x that the answer should be equal to 2.
check https://www.wolframalpha.com/input/?i=x%5Ex%3D2

mehdi bagheri
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  • How does this answer the question? – Michael Hoppe Nov 23 '20 at 19:45
  • @ChrisSteinbeckBell All this means is that the option of $x = 3/4$ comes closest to the exact value of the (presumably) transcendental value $x$. Assuming that the OP has not garbled the original problem, the OP is fully justified in presuming that one of the multiple choice options will exactly equal $x$. – user2661923 Nov 23 '20 at 19:46
  • @MichaelHoppe Totally agree to you. I think this answer doesn't really helps but rather is more like using a computer aided solution, which by the way produces $x=-2$ and $x=\frac{2}{3}$ as solutions if you go backwards using the official answer as a guide. – Chris Steinbeck Bell Nov 24 '20 at 01:24
  • @user2661923 Howdy. I'm not very familiar with transcendental numbers but I know the concept from wikipedia as mentioned in above comments. I copied it right from the source and the problem is as accurate as it is. But I'm starting to believe that there could be a typo. Assuming there was no typo. Do you confirm that $x=\frac{3}{4}$ is an approximation of the exact value?. If so, how did you obtained? – Chris Steinbeck Bell Nov 24 '20 at 01:26
  • @ChrisSteinbeckBell No, according to the equation $x^x = 2$ it is not an approximation. See the wolfram alpha link in mehdi bagheri's answer. – user2661923 Nov 24 '20 at 01:30
  • @user2661923 I did exactly that and used wolfram alpha's link but it did produced $x\approx 1.55961$. I disagree with the solution provided in this answer. If there isn't any error in my steps it should be $x^{-x}=\frac{1}{2}$. Sorry, I can't do more as I'm stuck. – Chris Steinbeck Bell Nov 24 '20 at 01:36
  • @ChrisSteinbeckBell $x^x = 2$ is equivalent to $x^{-x} = (1/2).$ I suggest that you email your teacher with an explanatory note (or somehow communicate with your teacher). Give your teacher the link to this webpage, and see what your teacher says. – user2661923 Nov 24 '20 at 01:39