Question 851151
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Hi,
{{{a^0 = 1}}}
{{{(a^p)^q = a^(p*q)}}}
{{{( x^m )( x^n ) = x^( m + n )}}}****{{{y^(mx)y^n = y^(mx+n)}}}
{{{x^m/x^n=x^(m-n)}}}
{{{(xy)^n = x^n*y^n}}}