Question 683635
<pre>
I think you just didn't remember that the square root is the same
as the {{{1/2}}} power.

{{{ log ( b, sqrt (32/27) ) }}} 

We observe that 32 = 2*16 = 2*2*8 = 2*2*2*4 = 2*2*2*2*2 = 2<sup>5</sup>
We observe that 27 = 3*9 = 3*3*3 = 3<sup>3</sup>

{{{ log ( b, sqrt (2^5/3^3) ) }}}

We observe that the square root is the {{{1/2}}} power:

{{{matrix(2,1,"",

 log ( b, (2^5/3^3)^(1/2) )) }}}

Now we use a rule of logarithms that allows us to move the {{{1/2}}}
exponent in front as a coefficient of the log:

{{{expr(1/2)log ( b, (2^5/3^3) ) }}}

Now we can write the log of a quotient as the difference of the logs,
being careful to place the difference of logs in parentheses so that
the {{{1/2}}} is multiplied by both terms:

{{{expr(1/2)(log ( b, 2^5 )-log(b,3^3) )}}}

Now on each of those terms inside the parentheses, we again use the
rule of logarithms that allows us to move the 5 and 2 exponents in 
front as a coefficient of the logs:

{{{expr(1/2)(5*log ( b, 2 )-3*log(b,3) )}}}

Finally we substuitute A for {{{log(b,2)}}} and C for {{{log(b,3)}}}

{{{expr(1/2)(5A-3C )}}}

Yoou can leave it like that or write it over 2 like this:

{{{(5A-3C )/2}}}

or as

{{{5A/2-3C/2}}}

or as

{{{expr(5/2)A-expr(3/2)C)}}}

Edwin</pre>