Why is 5-⅔ (5 raised to the power of -2/3) equal to 1/5⅔ (1 divided 5 raised to the power of 2/3)? I mean to say how does the negative sign of the exponent ⅔ change to positive when 5-⅔ (5 raised to the power of -2/3) is inversed?

Thanks,

Ron

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- Apr 24th 2010, 03:21 AMrn5aExponents
Why is 5-⅔ (5 raised to the power of -2/3) equal to 1/5⅔ (1 divided 5 raised to the power of 2/3)? I mean to say how does the negative sign of the exponent ⅔ change to positive when 5-⅔ (5 raised to the power of -2/3) is inversed?

Thanks,

Ron - Apr 24th 2010, 03:25 AMProve It
- Apr 24th 2010, 03:40 AMrn5a
- Apr 24th 2010, 03:44 AMBacterius
Because it is. You can explain it using this formula :

$\displaystyle \frac{a^m}{a^n} = a^{m - n}$

In our case, $\displaystyle m = 0$ because $\displaystyle a^0 = 1$, and thus it follows that :

$\displaystyle \frac{a^0}{a^n} = a^{0 - n}$

Which is equivalent to :

$\displaystyle \frac{1}{a^n} = a^{-n}$

Now, you are probably going to ask us why $\displaystyle \frac{a^m}{a^n} = a^{m - n}$. To this I will answer : "Why does 1 + 0 = 1 ?". :D - Apr 24th 2010, 04:05 AMrn5a
Thanks mate....that was a great explanation. BTW why is 1 + 0 = 1?http://www.mathhelpforum.com/math-he...ons/icon12.gif

- Apr 24th 2010, 04:25 AMBacterius
- Apr 24th 2010, 04:36 AMProve It
- Apr 24th 2010, 09:34 AMChokfull
I wouldn't say "It's just obvious". When you have $\displaystyle 5^2$, it means $\displaystyle 1*5*5$. When the exponent, which means

*the number of 5s you multiply the 1 by*is negative, you start dividing. Therefore, $\displaystyle 5^{-1}=\frac {1} {5}$ - Apr 24th 2010, 04:30 PMProve It