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

Guest Aug 27, 2014

#1**+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**+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:

Alan
Aug 27, 2014

#3**0 **

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

Melody
Aug 27, 2014

#4**+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**+5 **

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

Melody
May 11, 2015

#6**+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**+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