SOLUTION: Please explain why {{{(log(base b) sqrt(27))/3= log(base b) (sqrt(27)/3)}}}.

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Question 950771: Please explain why %28log%28base+b%29+sqrt%2827%29%29%2F3=+log%28base+b%29+%28sqrt%2827%29%2F3%29.
Found 2 solutions by josgarithmetic, Fombitz:
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
Nearest to be figured, your notation is wrong.

Underneath the rendering tags you show
(log(base b) sqrt(27))/3= log(base b) (sqrt(27)/3)

That text as you wrote seems to show the exact same number for the left member as for the right member. The parentheses on the left member are mishandled.

Your attempt appears to be asking about
log(b,sqrt(27)/3)=log(b,(sqrt(27)/3))

Which will appear more normally when rendering tagging added,
log%28b%2Csqrt%2827%29%2F3%29=log%28b%2C%28sqrt%2827%29%2F3%29%29
Which ARE THE SAME. There is some visual trouble on the left member because the denominator 3 is a little more difficult to interpret in the human inspection; which is why the doubled parentheses on the right member makes this easier to read. The input for both log expressions is sqrt%2827%29%2F3. The base, shown as a not-so-good subscript, is b.

Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
log%28b%2Csqrt%2827%29%29=%281%2F2%29log%28b%2C%2827%29%29
log%28b%2Csqrt%2827%29%29=%281%2F2%29log%28b%2C%283%5E3%29%29
log%28b%2Csqrt%2827%29%29=%283%2F2%29log%28b%2C%283%29%29
.
.
.
log%28b%2C%28sqrt%2827%29%2F3%29%29=log%28b%2C%28sqrt%2827%29%29%29-log%28b%2C%283%29%29
log%28b%2C%28sqrt%2827%29%2F3%29%29=%281%2F2%29log%28b%2C%2827%29%29-log%28b%2C%283%29%29
log%28b%2C%28sqrt%2827%29%2F3%29%29=%281%2F2%29log%28b%2C%283%5E3%29%29-log%28b%2C%283%29%29
log%28b%2C%28sqrt%2827%29%2F3%29%29=%283%2F2%29log%28b%2C%283%29%29-log%28b%2C%283%29%29
log%28b%2C%28sqrt%2827%29%2F3%29%29=%283%2F2-2%2F2%29log%28b%2C%283%29%29%29
log%28b%2C%28sqrt%2827%29%2F3%29%29=%281%2F2%29log%28b%2C%283%29%29
So then, if you divide the first term by 3, you'll get the second term.