How do you condense #2log(x-3)+log(x+2)-6logx#?

1 Answer
Bdub
Mar 22, 2016

#2log(x-3)+log(x+2)-6logx##=log(((x-3)^2 (x+2))/x^6)#

Explanation:

#2log(x-3)+log(x+2)-6logx#
#=log(x-3)^2+log(x+2)-logx^6#-> use property #log_bx^n=nlog_bx#
#=log(((x-3)^2 (x+2))/x^6)#->use properties #log_b(xy)=log_bx+log_by, and log_b(x/y) = log_b x-log_b y#

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