Exponential proof

**thedoctor818** · April 22nd 2010, 02:21 PM

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

Any help would eb appreciated.

**southprkfan1** · April 22nd 2010, 06:20 PM

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

**Drexel28** · April 22nd 2010, 06:22 PM

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.

**HallsofIvy** · April 23rd 2010, 03:34 AM

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.

**thedoctor818** · April 23rd 2010, 10:05 AM

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

**thedoctor818** · April 23rd 2010, 10:28 AM

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

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