Question

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.

Answer 1

exercise to the reader ..eh?

Answer 2

take log_x of both sides

Answer 3

idk why but I'm still uncomfortable with regular logs.. Always have to look up properties

Answer 4

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.

Answer 5

*lose

Answer 6

I have the proof copied, but I won't post it yet
it is easy though...

Answer 7

A corollary of this would be:
\[x ^{\log_{x}z }=z\]
Just take y=z

Answer 8

*y=x

Answer 9

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?

Answer 10

Como quieras

Answer 11

nah

Answer 12

now I just noticed yours is so much simpler anyway James

Answer 13

For the record then:
\[ \huge \log_y( x^{\log_y z}) = (\log_y x)(\log_yz) = \log_y(z^{\log_y x})\]

Answer 14

the chopped off term is just \[ \huge \log_y(z^{\log_y x}) \]

Answer 15

Taking log with base y would give just a 2-line proof
\[antilog(\log_{y} z \times \log_{y}x)=antilog(\log_{y}z)^{\log_{x}y } =z ^{\log_{y}x }\]

Answer 16

\[\Large {x^{\frac{\log_x z}{\log_x y}}} = {z^{\frac{\log_z x}{\log_z y}}}\] \[z - y = x - y \implies z = x\]

Answer 17

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?

Answer 18

I thought we were supposed to get the solution, my bad :/