My question is to prove the following:
$\displaystyle (a^{b1})^{b2}=a^{b1b2}.$
Any help would eb appreciated.
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.
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