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Question #55de4

1 Answer

Nov 14, 2016

Assuming you meant #log_a(a/sqrtx)-log_a(sqrtax)#

#1/2-3/2log_ax#

Explanation:

#log_a(a/sqrtx)-log_a(sqrtax)=log_a((a/sqrtx)/(sqrtax))=log_a(sqrta/(x sqrtx))=log_asqrta-log_ax^(3/2)=1/2-3/2log_ax#

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