# Thread: Calculate power of negative decimal number

1. ## Calculate power of negative decimal number

Hello
I am new in this forum
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

2. Originally Posted by shayw
Hello
I am new in this forum
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.

3. Looks good to me.

4. ## 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

5. 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?

6. The exponent of the root? I think you mean the exponent of the base.
Steven

7. 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.

$\displaystyle (-10)^{2.7} = (-10)^2 \cdot (-10)^{\frac7{10}} = \underbrace{100}_{real} \cdot \underbrace{\sqrt[10]{(-10)^7}}_{imaginary}$

8. Hi Bjhopper
If i understand the exponant must be a rationnal number
so -3^e or -3^pi can not be determined

Thanks

9. Yes, what I said about fractions was mostly wrong. I the power is irrational then the base can not be negative

10. Sorry, I did not know that yout notion was valid in another country

11. 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 $\displaystyle x$.

12. 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

13. 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.

14. Ok, that cleared everything up for me. Thanks!