Question

log X na base 2 + 4 log de 8 na base x = 8. me ajudem a responder pfvr

Answer 1

Propriedades:

$\mathsf{i) \:\log _b\left(a\right)=\dfrac{\log _c\left(a\right)}{\log _c\left(b\right)}}\\\\\\\mathsf{ii) \:\log _b\left(a\right)=c\:\:\:\:\:\Longleftrightarrow \:\:\:\:\:b^c=a}$
= = = = =

Dada a equação logaritmântica:

$\mathsf{\log _2\left(x\right)+4\log _x\left(8\right)=8}$

Mudando todos os logaritmos para uma base em comum, utilizando a propriedade (i):

$\mathsf{\log _2\left(x\right)+\dfrac{4\cdot \log _2\left(8\right)}{\log _2\left(x\right)}=8}\\\\\\\mathsf{\log _2\left(x\right)+\dfrac{4\cdot 3}{\log _2\left(x\right)}=8}\\\\\\\mathsf{\log _2\left(x\right)+\dfrac{12}{\log _2\left(x\right)}=8}\\\\\\\mathsf{}\\\\\\\mathsf{}\\\\\\\mathsf{}\\\\\\\mathsf{}\\\\\\\mathsf{}$

Para facilitar irei substituir $\mathsf{\log _2\left(x\right)=k}$:

$\mathsf{k+\dfrac{12}{k}=8}\\\\\\\mathsf{\dfrac{k\cdot k}{k}+\dfrac{12}{k}=8}\\\\\\\mathsf{\dfrac{k^2+12}{k}=8}\\\\\\\mathsf{k^2+12=8k}\\\\\\\mathsf{k^2-8k+12=0}$

Utilizando o método de resolução de equação do 2ª grau, onde:

$\boxed{\mathsf{a=1;\:b=-8;\:c=12}}$

$\mathsf{k=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}}\\\\\\\mathsf{k=\dfrac{-\left(-8\right)\pm \sqrt{8^2-\left(4\cdot 1\cdot 12\right)}}{2\cdot 1}}\\\\\\\mathsf{k=\dfrac{8\pm \sqrt{64-48}}{2}}\\\\\\\mathsf{k=\dfrac{8\pm \sqrt{16}}{2}}\\\\\\\mathsf{k=\dfrac{8\pm 4}{2}}\\\\\\\mathsf{k=4\pm 2}\\\\\\\mathsf{k_1=4+2\:\:\:\:\:\:\:\:\:ou\:\:\:\:\:\:\:\:\:k_2=4-2}\\\mathsf{k_1=6\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:ou\:\:\:\:\:\:\:\:\:k_2=2}$

Ou seja, o k pode assumir dois valores (6 ou 2). Retornando para a variavel inicial ($\mathsf{\log _2\left(x\right)}$):

๏ Primeira solução:

$\mathsf{k_1=\log _2\left(x\right)}\\\\\mathsf{6=\log _2\left(x\right)}\\\\\mathsf{2^6=x}\\\\\mathsf{x=64}$

๏ Segunda solução:

$\mathsf{k_2=\log _2\left(x\right)}\\\\\mathsf{2=\log _2\left(x\right)}\\\\\mathsf{2^2=x}\\\\\mathsf{x=4}$

Portanto a solução sera:

$\boxed{\mathsf{S=\left\{x\in \mathbb{R},\:x=4\:\:\:ou\:\:\:x=64\right\}}}\: \: \checkmark$