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

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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#