#4**+5 **

There's nothing wrong with your logic Melody; there are 3 possible solutions:

$${{\mathtt{a}}}^{{\mathtt{3}}} = {\left(-{\mathtt{3}}\right)}^{\left(-{\mathtt{4}}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{a}} = {\frac{\left({\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}\right)}\right)}}\\

{\mathtt{a}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}\right)}\right)}}\\

{\mathtt{a}} = {\frac{{\mathtt{1}}}{{{\mathtt{3}}}^{\left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}\right)}}}\\

\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{a}} = {\mathtt{\,-\,}}{\mathtt{0.115\: \!560\: \!212\: \!391\: \!772\: \!5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.200\: \!156\: \!159\: \!196\: \!130\: \!2}}{i}\\

{\mathtt{a}} = {\mathtt{\,-\,}}\left({\mathtt{0.115\: \!560\: \!212\: \!391\: \!772\: \!5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.200\: \!156\: \!159\: \!196\: \!130\: \!2}}{i}\right)\\

{\mathtt{a}} = {\mathtt{0.231\: \!120\: \!424\: \!783\: \!544\: \!9}}\\

\end{array} \right\}$$

.

Alan
Nov 28, 2014

#1**+5 **

Input was

(-3)^4/3

-----------------------------------------------------

I am actually surprised that the calc interpreted it this way.

I expected it to be interpreted as $$\frac{(-3)^{-4}}{3}$$ in which case the answer would be

$${\frac{\left({\left(-{\mathtt{3}}\right)}^{-{\mathtt{4}}}\right)}{{\mathtt{3}}}} = {\frac{{\mathtt{1}}}{{\mathtt{243}}}} = {\mathtt{0.004\: \!115\: \!226\: \!337\: \!448\: \!6}}$$

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Now lets go back to the calculator interpretation

$${\left(-{\mathtt{3}}\right)}^{{\mathtt{\,-\,}}{\frac{{\mathtt{4}}}{{\mathtt{3}}}}} = \underset{{\tiny{\text{Error: not real}}}}{{{\left(-{\mathtt{3}}\right)}}^{-{\mathtt{1.333\: \!333\: \!333\: \!333\: \!333\: \!3}}}}$$

If I do this by hand I could get

$$(-3)^{(-4/3)}\\\\

=\left[(-3)^{-4}\right]^{1/3}\\\\

=\left[\frac{1}{(-3)^{+4}}\right]^{1/3}\\\\

=\left[\frac{1}{-3*-3*-3*-3}\right]^{1/3}\\\\

=\left[\frac{1}{81}\right]^{1/3}\\\\

=\frac{1}{\sqrt[3]{81}}\\\\

\approx 0.2311$$

However I have tried it on 3 different calculators and none of them will give a 'real number' answer.

When you raise negative numbers to fractional powers it certainly cause problems.

Consider when the numerator is 1 and the denominator is an even number, like

$${\left(-{\mathtt{9}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{2}}}}\right)} = {\sqrt{-{\mathtt{9}}}}$$ there is definitely no **real number answer **to this.

$${\left(-{\mathtt{9}}\right)}^{\left({\frac{{\mathtt{1}}}{{\mathtt{4}}}}\right)}$$ there is definitely no **real number answer **to this either.

I am not giving you a proper answer here. I am just discusing your problem.

I think the calculators work these out with a generated series. Powers of neg numbers just do not work.

I am sure that Alan and maybe Heureka and CPhill can have a lot more to say on this subject.

----------------------------------------------------------------------------------------

Please note that (-8)^(-1/3) works just fine

$${\left(-{\mathtt{8}}\right)}^{{\mathtt{\,-\,}}\left({\frac{{\mathtt{1}}}{{\mathtt{3}}}}\right)} = \underset{{\tiny{\text{Error: not real}}}}{{{\left(-{\mathtt{8}}\right)}}^{-{\mathtt{0.333\: \!333\: \!333\: \!333\: \!333\: \!3}}}}$$

**Correction** - it works just fine on my CASIO - it does not work on the web2 calc

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**I know this has been discussed before but I'd really like other mathematicians to discuss this again.**

**Thank you.**

Melody
Nov 28, 2014

#2**+5 **

Calculators often use the following method to calculate fractional powers:

If we have a = b^{c} they take logs to get ln(a) = c*ln(b). Now raise both sides to the power e to get a = e^{c*ln(b)}.

If b is negative they can complain!

In fact: (-3)^{-4/3} = -0.11556 + 0.200156i

.

Alan
Nov 28, 2014

#3**0 **

I know that is right Alan but what is wrong with the logic on my hand done answer?

Also

Why does (-8)^(-1/3) give a real answer of -1/2 on some calcs (eg casio) and 'error' on other calcs?

Melody
Nov 28, 2014

#4**+5 **

Best Answer

There's nothing wrong with your logic Melody; there are 3 possible solutions:

$${{\mathtt{a}}}^{{\mathtt{3}}} = {\left(-{\mathtt{3}}\right)}^{\left(-{\mathtt{4}}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{a}} = {\frac{\left({\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,-\,}}{\mathtt{1}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}\right)}\right)}}\\

{\mathtt{a}} = {\mathtt{\,-\,}}{\frac{\left({\sqrt{{\mathtt{3}}}}{\mathtt{\,\times\,}}{i}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}{\left({\mathtt{2}}{\mathtt{\,\times\,}}{{\mathtt{3}}}^{\left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}\right)}\right)}}\\

{\mathtt{a}} = {\frac{{\mathtt{1}}}{{{\mathtt{3}}}^{\left({\frac{{\mathtt{4}}}{{\mathtt{3}}}}\right)}}}\\

\end{array} \right\} \Rightarrow \left\{ \begin{array}{l}{\mathtt{a}} = {\mathtt{\,-\,}}{\mathtt{0.115\: \!560\: \!212\: \!391\: \!772\: \!5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.200\: \!156\: \!159\: \!196\: \!130\: \!2}}{i}\\

{\mathtt{a}} = {\mathtt{\,-\,}}\left({\mathtt{0.115\: \!560\: \!212\: \!391\: \!772\: \!5}}{\mathtt{\,\small\textbf+\,}}{\mathtt{0.200\: \!156\: \!159\: \!196\: \!130\: \!2}}{i}\right)\\

{\mathtt{a}} = {\mathtt{0.231\: \!120\: \!424\: \!783\: \!544\: \!9}}\\

\end{array} \right\}$$

.

Alan
Nov 28, 2014