Solving Recursion Trees with Log Properties of a Constant

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Discussion Overview

The discussion centers around the properties of logarithms when applied to constants, particularly in the context of solving recursion tree problems. Participants explore the equivalence of expressions involving logarithms and powers, specifically comparing n^(log4 3) and 3^(log4 n).

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant seeks clarification on the log properties related to constants and their equivalence in the context of recursion trees.
  • Another participant explains the relationship between the expressions 4^x=3 and x=log4 3, demonstrating how to manipulate the expressions to show equivalence.
  • The manipulation involves substituting 3 with 4^(log4 3) and applying exponent properties to derive that 3^(log4 n) is equivalent to n^(log4 3).

Areas of Agreement / Disagreement

Participants appear to agree on the mathematical manipulations leading to the equivalence of the two expressions, though the initial question of their equivalence is posed without a definitive resolution.

Contextual Notes

The discussion does not resolve potential limitations or assumptions regarding the properties of logarithms or the specific context of recursion trees.

Who May Find This Useful

Readers interested in logarithmic properties, recursion tree analysis, or mathematical reasoning in algorithm complexity may find this discussion beneficial.

LauraSuh
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Hi, I'd like to know the log properties when they are the power of a constant. I've searched everywhere but I can't find it. the reason I want to know it is that, when solving a recursion tree problem, my teacher got an result of n^(log4 3) but I got 3^(log4 n). the base of the log haven't changed, but i'd like to know if those are equivalent and if possible, in case the answer is that they are equal, why.
Thx in advance!
 
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If we have to solve for x in 4^x=3 then we have defined a log function to answer this problem for us. We denote the answer to this problem as x=\log_43. This means that if we plug that expression for x back into the original problem, we get

4^{\log_43}=3 (1)


Now, for your answer
3^{\log_4n} (2)

notice that we can plug the expression I gave in equation (1) into equation (2) to replace the 3. i.e.

3^{\log_4n}=\left(4^{\log_43}\right)^{\log_4n}

and by exponent properties (a^b)^c=a^{bc}=a^{cb}=(a^c)^b we then have

=\left(4^{\log_4n}\right)^{\log_43}

But now notice the part in the brackets is equivalent to equation (1), that is, the log base 4 cancels the base of the exponent since they're the same value.

4^{\log_4n}=n

and so you finally have that your answer is equivalent to

n^{\log_43}
 
Thank you so much! You're like my super hero right now! =)
 
You're welcome :smile: By the way, may I ask what the question was? I'm guessing you were finding a time complexity, but I'd just like to see how that order arose.
 

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