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Answer 1 · 2016-11-05T02:29:33

#a=b and a=1/b^2#

Given #log_ab^2 = c and log_b a=c-1 #

So

#log_ab^2-1 = c-1= log_b a#

#=>log_ab^2-1 =1/log_a b#

#=>(2log_ab-1) log_a b=1#

#=>2(log_ab)^2- log_a b-1=0#

taking # log_a b=x# we get

#2x^2-x-1=0#

#=>2x^2-2x+x-1=0#

#=>2x(x-1)+1(x-1)=0#

#=>(x-1)(2x+1)=0#

So when #x-1=0 =>x=1#

#=>log_a b=1=log_a a#

#=>a=b#

Again when #2x+1=0#

#=>2log_a b+1=0#

#=>log_a b^2=-1#

#=>b^2=a^-1=1/a#

#=>a=1/b^2#