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

[ubergewehr273]

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.

 

[Fightfish]

A better way to do this is to recast the inequality as
$$3 > \log_{5}34 > 2$$
and rewrite $3$ and $2$ in terms of base-5 logarithms as well.

 

[ubergewehr273]

Thanks. Made it much easier than my solution.