## mitodoteira 2 years ago Prove: a^b*a^c=a^(b+c) for real a, b, c.

1. mitodoteira

$a,b,c \in \mathbb{R}$$Prove: a^{b}a^{c}=a^{b+c}$

2. Noura11

Wait ! a should be a positive real !

3. mitodoteira

a>1 I forgot to say that. Facepalm.

4. Noura11

It is true for all a>0. We can say : $\Large a^ba^c=e^{\ln a^b}e^{\ln a^c}\\ ~~~~~~~~~\Large =e^{b\ln a}e^{c\ln a}\\ ~~~~~~~~~\Large=e^{b\ln a+c\ln a}\\ ~~~~~~~~~~\Large=e^{(b+c)\ln a}\\ ~~~~~~~~~~\Large=e^{\ln a^{b+c}} \\ ~~~~~~~~~~\Large=a^{b+c}$

5. mitodoteira

I can't use logs. I'm doing analysis and I'm still proving the fundamentals of real powers from the field axioms.

6. Noura11

OK ! If b and c are natural integers you can prove it using induction !

7. mitodoteira

b, c are just real.

8. Noura11

@mitodoteira My last reply is the 1st step of the proof !

9. mitodoteira

How?

10. sauravshakya

I dont think we can use induction here... three variables

11. sauravshakya

Isnt it the property??? I am not sure if it can be proven. example : we know a*b=b*a because its a property what how to Prove it.

12. mitodoteira

No, it isn't a property.

13. swissgirl

hmmmmm how do you define the exponential function?

14. swissgirl

http://math.stackexchange.com/questions/435751/proving-the-product-rule-for-exponents-with-the-same-base Here is the link to this particular question I asked on MSE. Take a look at both proofs. The proof using Least Upper Bound is more analytical and a touch harder to follow but I think that is the proof you are looking for.