respectively which are positive integers
The solutions of 1 and 2 areFind
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If we look carefully at the problem, In 1, we would observe that for any value of x other than 0 , we have expression equal to 0.so as x approaches 0 , the expression must be very close to 0.Now , we have been asked b to the power a and a being 0 makes the expression =1, and we need not even evaluate b.