My question is to prove the following:

$\displaystyle (a^{b1})^{b2}=a^{b1b2}.$

Any help would eb appreciated.

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- Apr 22nd 2010, 02:21 PMthedoctor818Exponential proof
My question is to prove the following:

$\displaystyle (a^{b1})^{b2}=a^{b1b2}.$

Any help would eb appreciated. - Apr 22nd 2010, 06:20 PMsouthprkfan1
- Apr 22nd 2010, 06:22 PMDrexel28
- Apr 23rd 2010, 03:34 AMHallsofIvy
What definitions do you have to work with? Certainly, for $\displaystyle b_1$ and $\displaystyle b_2$ positive integers, you can think of "$\displaystyle a^b_1$" as meaning "a multiplied by itself $\displaystyle b_1$ times" so that $\displaystyle (a^{b_1})^b_1$ is "(a multiplied by itself $\displaystyle b_1$ times) multiplied by itself $\displaystyle b_2$ times" so that $\displaystyle (a^{b_1})^{b_2}$ is a multiplied by itself a total of $\displaystyle b_1b_2$ times. Then, the standard procedure is to

**define**define $\displaystyle a^b$ for b negative, a fraction, and an irrational number so that $\displaystyle (a^b_1)(a^b_2)= a^{b_1+ b_2}$, $\displaystyle (a^b_1)^b_2= a^{b_1b_2}$, and so that $\displaystyle a^b$ is a continuous function of b. - Apr 23rd 2010, 10:05 AMthedoctor818
Can I say the following:

$\displaystyle (a^{b_1})^{b_2}=a^{b_1\;b_2}.$

$\displaystyle =\exp[b_2\log(a^{b_1})].$

$\displaystyle =\exp[b_2b_1\log(a)].$

$\displaystyle =a^{b_2\;b_1}.$

$\displaystyle =a^{b_1\;b_2}.$

-Michael - Apr 23rd 2010, 10:28 AMthedoctor818
Can I say the following:

$\displaystyle

(a^{b_1})^{b_2}=a^{b_1\;b_2}.\\

=\exp[b_2\log(a^{b_1})].\\

=\exp[b_2\;b_1\log(a)].\\

=a^{b_2\;b_1}.\\

=a^{b_1\;b_2}.

$

-Michael