log⁡(〖2t〗^2-3t-6)+log⁡(2)=2 log⁡(2t).

Answer 1

(2t)^2 -2t -6

=4t^2 -2t -6

= (4t-6)(t+1) =2(2t-3)(t+1)

log(2(2t-3)(t+1))= log2 +log(2t-3) +log(t+1)

2log2t= 2log2 +2logt

log((2t)^2 -2t- 6) + log2 = 2log2t

log2 + log(2t-3) + log(t+1) + log2 = 2log2 + 2logt

log(2t-3)+log(t+1) = logt^2

log((2t-3)(t+1)/t^2)=0

(2t-3)(t+1)/t^2=1

2t^2-t =t^2

t^2-t=0

t(t-1)=0

t=0 or 1

but this did change the -3t to -2t in the original problem

to allow for simple factoring, but same basic approach otherwise

use properties of logs

logab=loga+logb

log1=0