## anonymous 3 years ago prove the rule:

1. anonymous

$a^{x}=\exp(xlna)$

2. anonymous

and $\log _{a}=\frac{ lnx }{ lna }$

3. anonymous

$e^{xlna}=(e^{lna})^x=(e^{\log_ea})^x=a^x$

4. anonymous

For the second, do you mean$\log_a x=\frac{lnx}{lna}$?

5. anonymous

$\large \log_a x= \log_{a^{\log_a e}}x^{\log_a e}=\ln x^{\log_a e}$

6. anonymous

$\large \ln x^{\log_a e} ln a= lnx$ $\large ( x^{\log_a e} + a)=x$

7. anonymous

$\large x^{\log_a e}=x-a$ $\large x^{\log_a e}=x^{log_x{(x-a)}}$ $\large {\log_a e}={log_x{x-a}}$

8. anonymous

@henpen , yes sorry i forgot the x

9. anonymous

I'm not quite sure where to go after that

10. anonymous

am not sure i follow ur steps.. how did you get all the those terms in the first part?

11. anonymous

|dw:1352237942749:dw|

12. anonymous

|dw:1352237963548:dw|

13. anonymous

Or...|dw:1352238104786:dw|

14. anonymous

my problem isnt that i cant see what you are writing, i dont understand where you derive them from..

15. anonymous

You could try Taylor expansion for the thing

16. anonymous

Oh, OK$\large log_k a/log_kb=\log_k(a-b)$$\large log_k a \cdot log_kb=\log_k(a+b)$$\large log_a b=\log_{a^c}b^c$ Are these clear?

17. anonymous

the first two.. last one i have never seen..

18. anonymous

It's fairly obvious if you think about it|dw:1352238416055:dw|

19. anonymous

ok

20. anonymous

i'll ask my prof tmw :)

21. anonymous

I don't think$x=(x-a)^{lna}$ is correct, just insert a=e and you get $x=x-e$ Probably I've messed up somewhere