Question

2 Resolva as equações

a) Log x-2 9=2


b)Log x+1 (x² + 7)=2






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Answer 1

a) $\log_{x-2}9=2$

$(x-2)^2=9$

$(x-2).(x-2)=9$

$x^2-4x+4=9$

$x^2-4x+4-9=0$

$x^2-4x-5=0$

$\triangle=(-4)^2-4.1.(-5)$

$\triangle=16+20$

$\triangle=36$

$x1=\frac{-(-4)+\sqrt{36} }{2.1}$

$x1=\frac{4+6}{2}=\frac{10}{2}=5$

$x2=\frac{-(-4)-\sqrt{36} }{2.1}$

$\frac{4-6}{2}=\frac{-2}{2}=-1$

S={-1, 5}

b)$\log_{x+1}(x^2+7)=2$

$(x+1)^2=x^2+7$

$(x+1).(x+1)=x^2+7$

$x^2+2x+1=x^2+7$

$x^2+2x-x^2=7-1$

$2x=6$

$x=\frac{6}{2}$

x=3

Answer 2

a)

$log_{x - 2}(9) = 2 \\ {(x - 2)}^{2} = 9 \\ x - 2 = ±\sqrt{9} \\ x - 2 = ±3$

$x-2=3\\x = 3 + 2 \\ \boxed{x = 5}$

$\boxed{x-2=-3} \\\boxed{ x=2-3} \\ \boxed{x=-1}$

b)

$log_{x + 1}( {x}^{2} + 7) = 2 \\ {(x + 1)}^{2} = {x}^{2} + 7 \\ {x}^{2} + 2x + 1 = {x}^{2} + 7 \\ 2x + 1 = 7$

$\boxed{2x = 7 - 1}\\ \boxed{2x = 6} \\\boxed{x = \frac{6}{2} } \\ \boxed{x = 3}$