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How do you solve y^2(y^-2)

When multiplying y ^2 by y^-2 do you add the exponents?  If so that would be zero correct?
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1 Answer

Yes, you are right, Nicholas; the rule for multiplying two exponential terms with the same base is:
yayb = ya+b
So, yes, you are right again:
y2y-2 = y2+(-2) = y0
Now what does y0 = ?  (Hint: it's not zero)


y^0 =1 only if y≠0.