Question

log (x+6) + log (x+5) =3 log5

Answer 1

Condição de existência dos logaritmos é:

x+6 > 0
x > - 6

x+5 > 0
x > -5

S = (x ∈ |R | x > - 5 ) 

$Log (x+6) + log (x+5) =3.log5\\\\log((x+6).(x+5))=log5^3\\\\log(x^2+5x+6x+35)=log125\\\\x^2+11x+35=125\\x^2+11x-90=0\\\\\delta=11^2-4.1.(-90)\\\delta=121+360\\\delta=481\\\\\sqrt{481}=21,94\\\\x=\frac{-11\pm21,94}{2.1}\\\\x=\frac{10,94}{2}\\\\\boxed{x=5,47}~serve\\\\x'=\frac{-11-21,94}{2}\\\\\boxed{x'=-16,47}~n\~ao~serve$

S = (5,47 )  

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