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How do you condense $log (π) + 2 log r$ $log(pi) + 2log r$ ?

1 Answer

Jun 1, 2016

$log (π) + 2 log (r) = log (π r^{2})$ $log(pi)+2log(r)=color(green)(log(pir^2)$

Explanation:

Remember
$XXX a \cdot log (b) = log (b^{a})$ $color(white)("XXX")a*log(b) = log(b^a)$
and
$XXX log (c) + log (d) = log (c \cdot d)$ $color(white)("XXX")log(c)+log(d)=log(c*d)$

Therefore
$log (π) + 2 log (r)$ $log(pi)+2log(r)$
$XXX = log (π) + log (r^{2})$ $color(white)("XXX")=log(pi)+log(r^2)$

$XXX = log (π r^{2})$ $color(white)("XXX")=log(pir^2)$

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