<PRE><B>Question 4212</B>
first change the base, so that all are base 3...i will not write the base anymore, to ease the readability:


{{{(log135log15)/(log3) - (log5log405)/(log3)}}}


Since log3 to base 3 is 1, we get {{{log135log15 - log5log405}}}


135 = 5*27
405 = 5*81
and 15 = 5*3, so we get:


{{{log(27*5)log(3*5) - log5log(81*5)}}}


(log27+log5)(log3+log5) - log5(log81+log5)


log27log3 + log27log5 + log3log5 + (log5)^2 - log81log5 - (log5)^2


Now, log27 to base 3 is 3
and log81 to base 3 is 4, so...


3log3 + 3log5 + log5 + (log5)^2 - 4log5 - (log5)^2


everything cancels except for the 3, so this is the answer!


jon
</PRE>