How do you solve $log_5 2 + 2 log_5t= log_5(3-t)$ ?

Question

1246 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-04T17:04:54

$t=-3/2 or t=1$

Use Property : $log_b(xy)=log_bx+log_by$ and $log_bx^n=nlog_bx$

$log_5 2 +2log_5 t =log_5(3-t)$

$log_5 2+log_5 t^2 = log_5(3-t)$

$log_5 2t^2= log_5(3-t)$

$2t^2 = 3-t$

$2t^2+t-3=0$

$(2t+3)(t-1)=0$

$t=-3/2 or t=1$

How do you solve log_5 2 + 2 log_5t= log_5(3-t)log_5 2 + 2 log_5t= log_5(3-t)?