Limit of a Logarithm with Different Bases \lim_(n->oo)((2^(log_3 n))/(3^(\log_2 n)))

opositor5t

opositor5t

Open question

2022-08-19

Limit of a Logarithm with Different Bases
We are to compute
lim n > 2 log 3 n 3 log 2 n
Clearly the bases are reversed between the logarithm and exponents, so I can't seem to find any logarithm or exponential properties that can help simplify this problem.

Answer & Explanation

prelatiuvq

prelatiuvq

Beginner2022-08-20Added 6 answers

Let us define the function f ( . ) as follow:
f ( n ) = 2 log 3 n 3 log 2 n .
Using the base change formula, we can easily transform f ( . ) into the following function: (I used log and exp to denote the natural logarithm and the exponential function, respectively)
f ( n ) = exp ( log 2 log n log 3 e log 3 log n log 2 e )

f ( n ) = exp ( C log n ) = n C ,
where C is a constant. C = log 2 log 3 e log 3 log 2 e = log 2 log 3 log 3 log 2
Since C is negative, i.e., C < 0, then:
lim n + f ( n ) = 0.
Leypoldon

Leypoldon

Beginner2022-08-21Added 8 answers

Hint: We have log 2 n log 3 n. So our fraction is ( 2 3 ) log 3 n

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