JamesJ
  • JamesJ
Logarithms can be surprising. Suppose \( x, y, z > 0 \). Then \[ x^{\log_y z} = z^{\log_y x} \] I'll leave the proof to you.
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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bahrom7893
  • bahrom7893
exercise to the reader ..eh?
TuringTest
  • TuringTest
take log_x of both sides
bahrom7893
  • bahrom7893
idk why but I'm still uncomfortable with regular logs.. Always have to look up properties

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JamesJ
  • JamesJ
log base y of both sides. It's not hard. But even though you can write it down in one or two lines, I'm still surprised how this result does not feel intuitive and the kind of thing you'd loose marks for in high school algebra.
JamesJ
  • JamesJ
*lose
TuringTest
  • TuringTest
I have the proof copied, but I won't post it yet it is easy though...
Mani_Jha
  • Mani_Jha
A corollary of this would be: \[x ^{\log_{x}z }=z\] Just take y=z
Mani_Jha
  • Mani_Jha
*y=x
TuringTest
  • TuringTest
I don't know about log base y, I think I did one ok with log base x of both sides should I post it?
JamesJ
  • JamesJ
Como quieras
TuringTest
  • TuringTest
nah
TuringTest
  • TuringTest
now I just noticed yours is so much simpler anyway James
JamesJ
  • JamesJ
For the record then: \[ \huge \log_y( x^{\log_y z}) = (\log_y x)(\log_yz) = \log_y(z^{\log_y x})\]
JamesJ
  • JamesJ
the chopped off term is just \[ \huge \log_y(z^{\log_y x}) \]
Mani_Jha
  • Mani_Jha
Taking log with base y would give just a 2-line proof \[antilog(\log_{y} z \times \log_{y}x)=antilog(\log_{y}z)^{\log_{x}y } =z ^{\log_{y}x }\]
anonymous
  • anonymous
\[\Large {x^{\frac{\log_x z}{\log_x y}}} = {z^{\frac{\log_z x}{\log_z y}}}\] \[z - y = x - y \implies z = x\]
Mani_Jha
  • Mani_Jha
Hey Ishaan, z=x will make the identity completely meaningless, because then the Left hand side and right hand side will be the one same thing. Are you sure what you did is correct?
anonymous
  • anonymous
I thought we were supposed to get the solution, my bad :/

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