Raising a negative number to a non-integer power

Is it possible to do this???

For instance, (-2)^(3/2) appears to give no solution as ((-2)^3)^(1/2) = (-8)^(1/2) which has no real solutions.

However, (-2)^(6/4) gives ((-2)^6)^(1/4) = 64^(1/4) = 2.828...

So although I've raised -2 to effectively the same power, only one gives an answer. Am I doing something wrong? Please help!!

Reply 1

JamesyB

12

Is

\sqrt[4]{-2^{6}}

the same as

\sqrt[2]{-2^{3}}

??

I'm not convinced you can raise the power like this.

Reply 2

jammiebreadman

OP

JamesyB

Is

\sqrt[4]{-2^{6}}

the same as

\sqrt[2]{-2^{3}}

??

I'm not convinced you can raise the power like this.

Well, 2^(6/4) and 2^(3/2) are the same, so I thought it would be the same with -2. Though that might be a totally rubbish assumption.

Reply 3

Simplicity

17

jammiebreadman

Is it possible to do this???

For instance, (-2)^(3/2) appears to give no solution as ((-2)^3)^(1/2) = (-8)^(1/2) which has no real solutions.

However, (-2)^(6/4) gives ((-2)^6)^(1/4) = 64^(1/4) = 2.828...

So although I've raised -2 to effectively the same power, only one gives an answer. Am I doing something wrong? Please help!!

Yes if you write the negative number in polar form. Than you can raise it to a lot of thing even non integer(you should really say real numbers as non integer is ambigious).

If you know what polar form is then you could probably do a general case say (-a)^n in polar form.

You could even extend it to the idea of ln(-a).

P.S. Note, I don't think the anwser would be unique.

Reply 4

sonofdot

8

Its generally quite difficult to define raising negative numbers to arbitrary powers. If you define a^x as

e^{x \ln a}

you can sort of see why

Reply 5

mpdebo

1

Recall e^(it) = cos(t) + i*sin(t), where i = sqrt(-1). Then e^(i*pi) = -1 and for example,
(-5)^(-0.2) = ( 5 * exp(i*pi) ) ^(-0.2) = 5^(-0.2) * exp(i*(-0.2)*pi) = 0.7248 * ( cos(-0.2*pi) + i*sin(-0.2*pi) )