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Question #037a0

1 Answer

Konstantinos Michailidis

Feb 28, 2016

It is #a=3^(1/4)#

Explanation:

Hence #a^b=b^a# and #b=9a# we have that

#a^(9a)=(9a)^a=>9a*loga=a*log9a=> a*(9loga-log9a)=0=>loga^9=log9a=> a^9=9a=>a^8=9=>a=9^(1/8)=3^(2/8)=> a=3^(1/4)#

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