Proving 1/3 < log_{34} 5 < 1/2

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SUMMARY

The inequality \( \frac{1}{3} < \log_{34} 5 < \frac{1}{2} \) is proven by transforming the logarithmic expression using the change of base formula \( \log_b a = \frac{1}{\log_a b} \). The discussion highlights that \( \log_{34} 5 \) can be expressed as \( \frac{1}{\log_5 34} \), which simplifies to \( \frac{1}{\log_5 17 + \log_5 2} \). By recasting the inequality as \( 3 > \log_{5} 34 > 2 \) and rewriting the bounds in terms of base-5 logarithms, the proof becomes more straightforward and eliminates unnecessary complexity.

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  • Familiarity with logarithmic identities such as \( \log mn = \log m + \log n \).
  • Basic knowledge of inequalities and their manipulation.
  • Proficiency in working with logarithms in different bases.
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ubergewehr273
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Homework Statement


Prove that ##{1/3} < log_{34} 5 < {1/ 2}##

Homework Equations


##log_b a = {1/ log_a b}##
##logmn = logm + logn##

The Attempt at a Solution


##log_{34} 5 = {1/ log_5 34}##
##= 1/(log_5 17 + log_5 2)##
##=1/(1 + log_5 3 + log_5 2 + something)##
##=1/(1 + log_5 6 + something)##
##=1/(2+something~else)##

something else ##<1##

Hence it is greater than 1/3 and smaller than 1/2.
But I think mathematically speaking, that ##something## isn't supposed to be there.
 
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A better way to do this is to recast the inequality as
3 &gt; \log_{5}34 &gt; 2
and rewrite ##3## and ##2## in terms of base-5 logarithms as well.
 
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Likes   Reactions: RUber
Thanks. Made it much easier than my solution.
 

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