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Comparing exponents in an equation. 
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#1
Nov213, 01:18 AM

P: 137

a^{n} + b^{n}/ a^{n1} + b^{n1} = (a+b)/2
can we say n=1 by comparing exponents? is there any other solution of it? 


#2
Nov213, 02:01 AM

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P: 13,093

$$ \frac{a^n + b^n}{a^{n1}+b^{n1}}=\frac{a+b}{2}$$ rearranging: $$a^n+b^n = ab^{n1} + ba^{n1}$$
By inspection, n=1 will satisfy the relation for all real a and b. Are there any other examples? ##n=2## ##a^2+b^2=2ab \Rightarrow (ab)^2=0## ... so the relation with n=2 describes a curve (a parabola). So it kinda depends on what counts as an answer. (The problem is underspecified.) 


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