Question 1114111
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Use {{{log(a,b) = 1/log(b,a)}}} to get both logarithms with the same base:<br>
{{{1/2log(4,x) = (1/2)(1/log(4,x)) = (1/2)log(x,4) = log(x,(4^(1/2))) = log(x,2)}}}
{{{1/log(3,x) = log(x,3)}}}<br>
So then<br>
{{{1/2log(4,x)+1/log(3,x) = log(x,2)+log(x,3) = log(x,(2*3)) = log(x,6)}}}