Question 1107601
In the 'better solution' given by another tutor,  the answer comes out incorrect.

log(a) / log(b)  does not equal log(a/b)   because log(a/b) = log(a)-log(b).

I still don't have a non-calculator answer, maybe tomorrow, its late.


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Non-calculator solution:
Comparing {{{ log(2,(3)) }}} vs {{{ log(3,(5)) }}} is the same as comparing x and y in:
{{{ 2^x = 3 }}} vs {{{ 3^y = 5 }}}

Claim:  x>3/2
Set x=3/2 to check:

{{{  matrix(3,3, "" ,"", "",  2^(3/2), " = ", 8^(1/2), "", "" ,"") }}}
Comparing the square of this to {{{ 3^2 }}} we get  {{{ 8 < 9 }}} implying  x > 3/2  (using 3/2 as the exponent gives a number that is slightly too small) 


What if y = 3/2?

{{{ matrix(3,3, "", "", "" ,  3^(3/2), " = ", 27^(1/2), "" ,"", "") }}} 
Comparing the square of this to {{{ 5^2  }}} we get {{{ 27 > 25 }}} implying  y < 3/2 (using 3/2 as the exponent gives a number that is slightly too big)


We now have  {{{ y < (3/2) < x}}}    so we can say  {{{ y < x }}}.    
Therefore {{{ log(3,(5)) < log(2,(3)) }}} 
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