How do you condense #5 log_ (b) x + 6 log _(b) y#?

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Answer 1 · 2016-07-27T17:17:48

#log_b (x^5y^6)#

Since #nlog_b a=log_b a^n#,

then

#5log_b x+6log_b y=log_bx^5+log_by^6#

Since

#logx+logy=log(xy)#,

then

#log_bx^5+log_by^6=log_b (x^5y^6)#