Question 1035632
Try completely factorizing the starting expression.


Given:  {{{81^x}}}
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{{{(9*9)^x}}}-----------choice f
{{{(9^2)^x}}}
{{{9^(2x)}}}
{{{9^(x*2)}}}
{{{(9^x)(9^x)}}}--------choice c


A certain rule of exponents is used, which goes like  {{{(b^x)^y=b^(xy)}}}.