# Exponential proof

• Apr 22nd 2010, 02:21 PM
thedoctor818
Exponential proof
My question is to prove the following:
$\displaystyle (a^{b1})^{b2}=a^{b1b2}.$

Any help would eb appreciated.
• Apr 22nd 2010, 06:20 PM
southprkfan1
Quote:

Originally Posted by thedoctor818
My question is to prove the following:
$\displaystyle (a^{b1})^{b2}=a^{b1b2}.$

Any help would eb appreciated.

$\displaystyle (a^{b1})^{b2} = e^{b2\ln{a^{b1}}} = e^{b2(\ln{e^{b1\ln{a}}})}$

But, $\displaystyle \ln{e^{b1\ln{a}}} = b1\ln{a}$

so, $\displaystyle (a^{b1})^{b2} = e^{b2(\ln{e^{b1\ln{a}}})} = e^{b1b2(\ln{a})} = a^{b1b2}.$
• Apr 22nd 2010, 06:22 PM
Drexel28
Quote:

Originally Posted by southprkfan1
$\displaystyle (a^{b1})^{b2} = e^{b2\ln{a^{b1}}} = e^{b2(\ln{e^{b1\ln{a}}})}$

But, $\displaystyle \ln{e^{b1\ln{a}}}) = b2b1\ln{a}$

so, $\displaystyle (a^{b1})^{b2} = e^{b2(\ln{e^{b1\ln{a}}})} = e^{b1b2(\ln{a})} = a^{b1b2}.$

Surely if the poster is asking the question they are at a point in their studies where they aren't supposed to use the natural logarithm. But, I guess without stating how or why he wants this proved your method is as good as any.
• Apr 23rd 2010, 03:34 AM
HallsofIvy
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 AM
thedoctor818
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 AM
thedoctor818
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