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

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:

(-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.   

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

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!

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

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? 

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?