How do you solve the equation #log_b8+3log_bn=3log_b(x-1)# for n?

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Answer 1 · 2016-12-04T09:40:30

#n=(x-1)/2#

#log_b 8+3log_b n=3log_b(x-1)#

Now as #alog_m b=log_m b^a# and #log_m x+log_m y=log_m xy#,

we can simplify above as under.

#log_b 8+3log_b n=3log_b(x-1)#

or #log_b 8+log_b n^3=log_b(x-1)^3#

or #log_b (8n^3)=log_b(x-1)^3#

or #8n^3=(x-1)^3#

and #n^3=(x-1)^3/2^3#

i.e. #n=(x-1)/2#