anonymous
  • anonymous
prove the rule:
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
\[a^{x}=\exp(xlna)\]
anonymous
  • anonymous
and \[\log _{a}=\frac{ lnx }{ lna }\]
anonymous
  • anonymous
\[e^{xlna}=(e^{lna})^x=(e^{\log_ea})^x=a^x\]

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

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