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can i calculate a negative number to the power of a fraction? ie (-1.42)^(1/7)

math
Guest Aug 27, 2014

Best Answer 

 #2
avatar+26322 
+10

There are 7 numbers which, when raised to the power of 7 result in -1.42.  However, all but one of them are complex.  Melody has identified the only real number solution.  Here are all the solutions:

7th root solutions

Alan  Aug 27, 2014
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7+0 Answers

 #1
avatar+90988 
+10

(-1.42)^(1/7)

 

This is the web2 calculator ouput

$${\left(-{\mathtt{1.42}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{7}}}}\right)} = \underset{{\tiny{\text{Error: not real}}}}{{{\left(-{\mathtt{1.42}}\right)}}^{{\mathtt{0.142\: \!857\: \!142\: \!857\: \!142\: \!9}}}}$$

 

However, this is the answer (approx because it is an irrational number)

$${\mathtt{\,-\,}}\left({\left({\mathtt{1.42}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{7}}}}\right)}\right) = -{\mathtt{1.051\: \!369\: \!751\: \!024\: \!905\: \!2}}$$

 

check:

$${\left(-{\mathtt{1.051\: \!369\: \!751\: \!029\: \!052}}\right)}^{{\mathtt{7}}} = {\mathtt{\,-\,}}{\frac{{\mathtt{71}}}{{\mathtt{50}}}} = -{\mathtt{1.420\: \!000\: \!000\: \!039\: \!206\: \!4}}$$

 

 

I think that this can only work if the denominator of the power is an odd number.

Most calculators have a real problems with this - sometimes you really do have to use your own brain. 

 

I invite further comment on this one.   

Melody  Aug 27, 2014
 #2
avatar+26322 
+10
Best Answer

There are 7 numbers which, when raised to the power of 7 result in -1.42.  However, all but one of them are complex.  Melody has identified the only real number solution.  Here are all the solutions:

7th root solutions

Alan  Aug 27, 2014
 #3
avatar+90988 
0

This is true Alan but very few people on this forum have heard of complex numbers.   

Melody  Aug 27, 2014
 #4
avatar+109 
+4

I really like the way how Melody is able to get a solution out of this question. 

 

But in my opinion, it is a irrational number as a result. Or, if you "don't know" these irrational numbers, it is simply not defined. Why? Because of this:

$${\left(-{\mathtt{1.42}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{7}}}}\right)} = {\sqrt[{{\mathtt{{\mathtt{7}}}}}]{-{\mathtt{1.42}}}}$$

 

And the result of any root with a negative radicant is either not defined or a irrational number. 

 

(I am pretty sure that you can not pull the negative sign out of the brackets. Even if there is a nice result ;) )

 

 

Always ways up for comments!

xerxes  May 11, 2015
 #5
avatar+90988 
+5

What I did was perfectly valid Xerxes,  even Alan got the same real answer as I did.  

Melody  May 11, 2015
 #6
avatar+109 
+4

I know, I realised that. Another thing I leant today - however, I am pretty sure they taught me that you can't get a root if the radicant is negative.

it would just work with odd denominators, wouldn't it? 

xerxes  May 11, 2015
 #7
avatar+90988 
+5


Yes xerxes that sounds right.  

Of course it would work for any but there would only be a real answer if the denominator was odd.

 

You might find this interesting Xerxes,

http://www.wolframalpha.com/input/?i=%28-5%29%5E%281%2F3%29

 

It shows you how to get the cubed root of -5.

There are 3 answers because it is a cubed root

The first one is -(5^1/3) which his approx -1.7  this is the only real one.

Can you work out how to get the other 2 just from looking at the Wolfram|Alpha complex number plane diagram?

Melody  May 12, 2015

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