Question #c4bbb

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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 $2x+1=0$

$=>2log_a b+1=0$

$=>log_a b^2=-1$

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

$=>a=1/b^2$