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

1 Answer

Konstantinos Michailidis

Apr 15, 2016

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#

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