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The question is:

Let $a>0$. Prove that for any numbers $x_1$ and $x_2$,

a. $a^{x_1} * a^{x_2} = a^{x_1 + x_2}$

b. $(a^{x_1})^{x_2} = a^{x_1*x_2}$

I know that I am supposed to somehow use the fact that $a^x=g(x\ln a)$. The options for proof are very limited because $x_1$ and $x_2$ can be any number, not just rational numbers.

luke
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user107853
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  • How have you defined the operation of exponentiation? The answer to your question might be very easy or very hard, depending on which definition you're working with. – Clive Newstead Jan 09 '14 at 21:24
  • Sorry yes it should be a^(x_1) * a^(x_2), I will fix that. The definition of a^x I am working with is a^x = g(xlna) where g:R-->R is the inverse of the natural logarithm. a>0 was given in the question. – user107853 Jan 09 '14 at 21:33
  • I don't see any place except on the sidebar where this is linked, but this is what you need. – John Jan 10 '14 at 17:26

1 Answers1

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See this answer: https://math.stackexchange.com/a/633853/17622 for part a. In the comments section you can see that the proof assumes Part B of your problem is true.

It remains to prove part b given your information: $a^x=e^{x\cdot \ln(a)}$.

Let $a,b,c \in \mathbb{R}$ with $a$ nonzero. We want to show that $(a^b)^c=a^{b\cdot c}$. Start with the LHS:

$(a^b)^c=e^{c\cdot\ln(a^b)}=e^{c\cdot b\cdot \ln(a)}=e^{\ln(a)\cdot (c\cdot b)}={a}^{c\cdot b}$

Community
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Christian Chapman
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