Question

o valor de x para que log2*, log2

(3x+2), log2 (10x+12) formem, nessa ordem, uma p.a

Answer 1

$P.A~(a_{1},a_{2},a_{3})\\\\r=a_{2}-a_{1}\\r=a_{3}-a_{2}\\\\r=r\\a_{2}-a_{1}=a_{3}-a_{2}\\a_{2}+a_{2}=a_{3}+a_{1}\\2a_{2}=a_{1}+a_{3}$
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$a_{1}=log_{2}(x)\\a_{2}=log_{2}(3x+2)\\a_{3}=log_{2}(10x+12)$


$2a_{2}=a_{1}+a_{3}\\2*log_{2}(3x+2)=log_{2}(x)+log_{2}(10x+12)\\log_{2}(3x+2)^{2}=log_{2}[x*(10x+12)]$

Retirando log dos 2 lados:

$(3x+2)^{2}=x(10x+12)\\(3x)^{2}+2*3x*2+2^{2}=10x^{2}+12x\\9x^{2}+12x+4=10x^{2}+12x\\9x^{2}+4=10x^{2}\\4=10x^{2}-9x^{2}\\4=x^{2}\\\sqrt{4}=x\\x=2$