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

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Answer 1 · 2016-04-15T00:33:31

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 $s>0$

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