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how come the calculator says that (-3)^-4/3 is not real?

Guest Nov 28, 2014

Best Answer 

 #4
avatar+26405 
+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
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5+0 Answers

 #1
avatar+91469 
+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     

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

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

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

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

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
avatar+26405 
+5

Calculators often use the following method to calculate fractional powers:

 

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

If b is negative they can complain!

 

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

.

Alan  Nov 28, 2014
 #3
avatar+91469 
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
avatar+26405 
+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
 #5
avatar+91469 
0

Thanks Alan    

Melody  Nov 28, 2014

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