# Calculate power of negative decimal number

• May 11th 2011, 10:45 AM
shayw
Calculate power of negative decimal number
Hello
I am new in this forum (Happy)
I try to calculate

{-10}^{ 2.6}
using the calculator I get an error
but i write that way
{-10}^{ 2.6} = {-10}^{ 2+ 6/10} = {-10}^{ 2+ 3/5} = {-10}^{2 } * {-10}^{3/5 }
= 100 * {{-10}^{ 3} }^{1/ 5} = 100*{-1000}^{ 1/5} \approx -398.107
i don't know if it's correct
so for {-10}^{ 2.8} i get 630.957344480193
• May 11th 2011, 11:20 AM
earboth
Quote:

Originally Posted by shayw
Hello
I am new in this forum (Happy)
I try to calculate

{-10}^{ 2.6}
using the calculator I get an error
but i write that way
{-10}^{ 2.6} = {-10}^{ 2+ 6/10} = {-10}^{ 2+ 3/5} = {-10}^{2 } * {-10}^{3/5 }
= 100 * {{-10}^{ 3} }^{1/ 5} = 100*{-1000}^{ 1/5} \approx -398.107
i don't know if it's correct
so for {-10}^{ 2.8} i get 630.957344480193

If the exponent of the root is odd you don't get a real result.

EDIT: Changed one word in the last sentence so it is (better?) understandable.
• May 11th 2011, 11:22 AM
DrSteve
Looks good to me.
• May 11th 2011, 05:03 PM
bjhopper
calculate power of negative decimal number
Hi shayw,
I assume you mean the power of a negative integer. You can simplify calculations by factoring the negative integer

-10^2.8 =(-1)^2.8 * 10^2.8
10^2.8 = 630.957
(-1)^2.8 =(-1)^2 *(-1)^.8=1*1 answer is positive

If the original problem was(-10)^2.7

(-10)^2.7 =(-1)^2.7 * 10^2.7
= (-1)^2 *(-1)^.5 * (-1)^.2 * 10^2.7
= 1*-1*1=-1

bjh
• May 11th 2011, 05:36 PM
mathprofessor
Quote:

Originally Posted by earboth

If the exponent of the root isn't odd you don't get a real result.

How can you tell if a decimal number is even or odd? Do you just look at the last digit? Please realize that 10.7 also equals 10.70 and 10.700, etc. Note that 70 (and 700, 7000, ...) is even, so is 10.70 now even?
• May 11th 2011, 06:01 PM
mathprofessor
The exponent of the root? I think you mean the exponent of the base.
Steven
• May 11th 2011, 10:00 PM
earboth
Quote:

Originally Posted by mathprofessor
How can you tell if a decimal number is even or odd? Do you just look at the last digit? Please realize that 10.7 also equals 10.70 and 10.700, etc. Note that 70 (and 700, 7000, ...) is even, so is 10.70 now even?

1. You know that

$\displaystyle r^{\frac ew} = \sqrt[w]{r^e}$

(In German the number w is called exponent of the root. Maybe that's the reason why I used a wrong expression)

2. Re-write an exponent as a completely simplified fraction. Then the denominator of this fraction indicates if it is an odd or even root.

$\displaystyle (-10)^{2.7} = (-10)^2 \cdot (-10)^{\frac7{10}} = \underbrace{100}_{real} \cdot \underbrace{\sqrt[10]{(-10)^7}}_{imaginary}$
• May 12th 2011, 11:01 PM
shayw
Hi Bjhopper
If i understand the exponant must be a rationnal number
so -3^e or -3^pi can not be determined

Thanks
• May 13th 2011, 11:52 AM
mathprofessor
Yes, what I said about fractions was mostly wrong. I the power is irrational then the base can not be negative
• May 13th 2011, 11:53 AM
mathprofessor
Sorry, I did not know that yout notion was valid in another country
• May 13th 2011, 11:58 AM
DrSteve
We can define exponentiation with irrational exponents by taking a limit.

$\displaystyle a^x=\lim_{n \to \infty} a^{x_n}$ where $\displaystyle (x_n)$ is a sequence of rationals converging to $\displaystyle x$.
• May 13th 2011, 12:51 PM
mathprofessor
Hi,
Yes, I know what you are saying. I clearly remember my teacher showing the sequence that approached sqrt 2. I do not see (yet) why such a sequence would help us know if for example (-3)^sqrt2 is a real number.
Thanks,
Steven
• May 14th 2011, 04:36 AM
HallsofIvy
Dr. Steve didn't say it would help determine if such a root was real or not- only that it could be used to find the root if it existed. If $\displaystyle \{r_n= \frac{a_n}{b_n}\}$ is a sequence of rational numbers converging to $\displaystyle \sqrt{2}$ then for any positive real number, X, $\displaystyle (X)^{\sqrt{2}}$ is defined as the limit of the sequence $\displaystyle X^{r_n}= \sqrt[b_n]{X^{a_n}}$. In general, negative real numbers to an irrational power are not defined.
• May 14th 2011, 09:22 AM
mathprofessor
Ok, that cleared everything up for me. Thanks!