# Exponential proof

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

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

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

Any help would eb appreciated.

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

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

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

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

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

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

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.
• April 23rd 2010, 03:34 AM
HallsofIvy
What definitions do you have to work with? Certainly, for $b_1$ and $b_2$ positive integers, you can think of " $a^b_1$" as meaning "a multiplied by itself $b_1$ times" so that $(a^{b_1})^b_1$ is "(a multiplied by itself $b_1$ times) multiplied by itself $b_2$ times" so that $(a^{b_1})^{b_2}$ is a multiplied by itself a total of $b_1b_2$ times. Then, the standard procedure is to define define $a^b$ for b negative, a fraction, and an irrational number so that $(a^b_1)(a^b_2)= a^{b_1+ b_2}$, $(a^b_1)^b_2= a^{b_1b_2}$, and so that $a^b$ is a continuous function of b.
• April 23rd 2010, 10:05 AM
thedoctor818
Can I say the following:

$(a^{b_1})^{b_2}=a^{b_1\;b_2}.$
$=\exp[b_2\log(a^{b_1})].$
$=\exp[b_2b_1\log(a)].$
$=a^{b_2\;b_1}.$
$=a^{b_1\;b_2}.$

-Michael
• April 23rd 2010, 10:28 AM
thedoctor818
Can I say the following:

$
(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