Question

log(x^2+x -6) na base 5 = log(3x + 2) na base 5 .Me ajudem

Answer 1

Se trata de uma equação logaritmica, como as bases são iguais basta igualar os logaritmandos:

x² + x - 6 = 3x + 2
x² - 2x - 8 =0

resolvendo encontrará delta 36 e as raízes 4 e -2, ver qual das duas atende a condição de existencia do logaritmo que é:

base = maior que zero e diferente de 1
logaritmando = maior que zero

base = ok
logaritmando = será o 4 a resposta.

Answer 2

Dada a propriedade:

๏ $\mathsf{\log _b\left(a\right)=\log _b\left(c\right)\:\:\:\:\Leftrightarrow \:\:\:\:a=c}$

Logo:

$\mathsf{\log _5\left(x^2+x-6\right)=\log _5\left(3x+2\right)}\\\mathsf{x^2+x-6=3x+2}\\\mathsf{x^2+x-6-3x-2=0}\\\mathsf{x^2-2x-8=0}\\\\\\\mathsf{\boxed{\mathsf{a=1;\:b=-2;\:c=-8}}}\\\\\\\\\mathsf{x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}}\\\\\\\mathsf{x=\dfrac{-\left(2\right)\pm \sqrt{2^2-\left[4\cdot 1\cdot \left(-8\right)\right]}}{2\cdot 1}}\\\\\\\mathsf{x=\drac{2\pm \sqrt{4+32}}{2}}\\\\\\\mathsf{x=\dfrac{2\pm \sqrt{36}}{2}}\\\\\\\mathsf{x=\dfrac{2\pm 6}{2}}\\\\\\\mathsf{x=1\pm 3}$

$\mathsf{\hookrightarrow x_1=1+3=4}\\\mathsf{\hookrightarrow x_2=1-3=-2}$

Testando as soluções:

◎ x = 4 (Verdadeiro)

$\mathsf{\log _5\left(4^2+4-6\right)=\log _5\left((3 \cdot 4)+2\right)}\\\mathsf{\log _5\left(16+4-6\right)=\log _5\left(12+2\right)}\\\mathsf{\log _5\left(14\right)=\log _5\left(14\right)}$

◎ x = -2 (Falso, pois o logaritmando deve ser maior que zero)

$\mathsf{\log _5\left(\left(-2\right)^2+\left(-2\right)-6\right)=\log _5\left(\left[3\cdot \:\left(-2\right)\right]+2\right)}\\\mathsf{\log _5\left(4-2-6\right)=\log _5\left(-6+2\right)}\\\mathsf{\log _5\left(-4\right)=\log _5\left(-4\right)}$

$\boxed{\mathsf{Resposta:\:S=\left\{x\in \mathbb{R}\:,\:x=4\right\}}}\: \: \checkmark$