How do you solve $log_7(2x^2 - 1) = log_7 4x$ ?

Question

2190 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-11-24T16:06:42

$2+sqrt6$

Note that if $log_ab=log_ac,$ then $b=c$ .

Because of this, we can say that $2x^2-1=4x$ .

$2x^2-4x-1=0$

$x=(4+-sqrt(16+8))/2$

$x=2+-sqrt6$

However, the only answer is the positive one, $2+sqrt6$ , sincein $log_ab$ , $b>0$ .

How do you solve log_7(2x^2 - 1) = log_7 4xlog_7(2x^2 - 1) = log_7 4x?