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Mathematics
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\[a^{x}=\exp(xlna)\]
and \[\log _{a}=\frac{ lnx }{ lna }\]
\[e^{xlna}=(e^{lna})^x=(e^{\log_ea})^x=a^x\]

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Other answers:

For the second, do you mean\[\log_a x=\frac{lnx}{lna}\]?
\[\large \log_a x= \log_{a^{\log_a e}}x^{\log_a e}=\ln x^{\log_a e}\]
\[\large \ln x^{\log_a e} ln a= lnx\] \[\large ( x^{\log_a e} + a)=x\]
\[\large x^{\log_a e}=x-a\] \[\large x^{\log_a e}=x^{log_x{(x-a)}}\] \[\large {\log_a e}={log_x{x-a}}\]
@henpen , yes sorry i forgot the x
I'm not quite sure where to go after that
am not sure i follow ur steps.. how did you get all the those terms in the first part?
|dw:1352237942749:dw|
|dw:1352237963548:dw|
Or...|dw:1352238104786:dw|
my problem isnt that i cant see what you are writing, i dont understand where you derive them from..
You could try Taylor expansion for the thing
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?
the first two.. last one i have never seen..
It's fairly obvious if you think about it|dw:1352238416055:dw|
ok
i'll ask my prof tmw :)
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

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