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

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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)

Comments

y^0 =1 only if y≠0.