Spencer’s favorite logarithm property! - Videos

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• Dude you are super cool

• There is a shorter proof. Just take log_b of both sides of equation
  log_b(a^log_b(c))=log_b(c^log_b(a))
  then use property of log of power (power becomes multiplier of log function)
  log_b(c)*log_b(a)=log_b(a)*log_b(c)
  and this is an obvious identity

• Very cool !! :) Hey, have you ever thought about using a smaller microphone?

• Sorry, but this super-long proof is just ridiculous[ly stupid]!! [EDIT: maybe stupid is too much..]
  Let’s rephrase the problem and try solving it in the comments:
  Goal: “We can switch ‘A’ and ‘C’ in ‘A^log_B(C)’”
  Proof: Hmm… Let’s see what’s going on here… We have a log base B in the exponent, our base is A though, let’s fix that!
  A^log_B(C) = (B^log_B(A))^log_B(C) = B^(log_B(A)*log_B(C))
  —oh… and we’re done, the ‘*’ is symmetric!

• Does anyone know what's the name of the song at the end of the video? It sounds amazing

• You could've proved this in two lines. Take log[B] of both sides of the original equation, and apply the exponent property of logarithms. By the commutative property of multiplication the expression on the LHS = the expression on the RHS. QED

• Alternative proof: just take log_b of both sides and take out the powers (2 line proof). Also, conditions needed are just A, B, C > 0 and B not 1. Nice video though.

• My favorite is for what value of x does in(log(x))=log(in(x))

• There is a much simpler explanation. Just take the log_B of both sides to get: log_B(C)*log_B(A) = log_B(A)*log_B(C). One or two steps, and DONE. Also I believe the actual restriction is A,B,C > 0 and B != 1. I consider it bad form to not give precise conditions.

• satisfying property !!