Question

How do you solve for s in `log_5(2s +3) = 2log_5(s)`?

Answer 1

We have that

#log_5(2s +3) = 2log_5(s)=>(2s+3)=s^2=> s^2-2s-3=0=>(s^2-1)-2s-2=0=> (s-1)*(s+1)-2(s+1)=0=> (s+1)*(s-1-2)=0=> (s+1)*(s-3)=0#

Finally `s=3` hence it should be `s>0`