Question 87117
Is this what you mean?

{{{log(10,sqrt((s-a)*(s-g))^2) }}} 


If so, then writing that logarithmic expression was the hard part of the problem!  Remember that squaring "undoes" the square root, so you might think that 
{{{sqrt(x^2) = x}}}.  However, this is not technically correct, since by the principle value definition, a square root cannot equal a negative number.  We therefore make the definition that 
{{{sqrt(x^2) = abs(x) }}}
{{{sqrt((s-a)*(s-g))^2 }}}= {{{abs((s-a)*(s-g)) }}}


By this definition, you can easily write this expression:
{{{log(10,sqrt((s-a)*(s-g))^2) }}}= {{{log(10,abs((s-a)*(s-g))) }}}


R^2 Retired from SCC