Question

exercício de equação logarítmica: 1+ log(x-4) na base 2/ log (raiz de x+3 - raiz de x-3) na base raiz de dois = 1

Answer 1

Olá Jordana.


Checando a condição de existência dos logaritmandos:


$\mathsf{x\begin{cases}\mathsf{x-4>0~\Rightarrow~x>4}\\\mathsf{x-3>0~\Rightarrow~x>3}\\\mathsf{x+3>0~\Rightarrow x>-3}\end{cases}}\\\\\\\mathsf{C.E~\{x\in\mathbb{R}:x>4\}}$


Organizando e resolvendo a equação:


$\mathsf{\dfrac{1+\ell og_2(x-4)}{\ell og{_\sqrt{2}}(\sqrt{x+3}-\sqrt{x-3})}=1}}\\\\\\\mathsf{\ell og_2(2)+\ell og_2(x-4)=\ell og_{\sqrt{2}}(\sqrt{x+3}-\sqrt{x-3})}\\\\\underline{\qquad\qquad\qquad\qquad}\\\\\mathsf{Mudan\c{c}a~de~base:}\\\\\\\mathsf{\ell og_{\sqrt{2}}(\sqrt{x+3}-\sqrt{x-3})=\dfrac{\ell og_2(\sqrt{x+3}-\sqrt{x-3})}{\ell og_2(\sqrt{2})}}\\\\\mathsf{\underline{\qquad\qquad\qquad\qquad}}\\\\\\\mathsf{\ell og_2(2\cdot[x-4])=\dfrac{\ell og_2(\sqrt{x+3}-\sqrt{x-3})}{\ell og_2(\sqrt{2})}}$

$\mathsf{\ell og_2(2\cdot[x-4])=\dfrac{\ell og_2(\sqrt{x+3}-\sqrt{x-3})}{\ell og_2(\sqrt{2})}}\\\\\\\mathsf{\ell og_2(2x-8)=\dfrac{\ell og_2(\sqrt{x+3}-\sqrt{x-3)}}{\dfrac{1}{2}}}\\\\\\\mathsf{\ell og_2(2x-8)=\ell og_2(\sqrt{x+3}-\sqrt{x-3)}\cdot\dfrac{2}{1}}\\\\\\\mathsf{\ell og_2(2x-8)=\ell og_2([\sqrt{x+3}-\sqrt{x-3}]^2)}\\\\\\\mathsf{2x-8=(\sqrt{x+3}-\sqrt{x-3})^2}\\\\\mathsf{2x-8=x+3-2\cdot\sqrt{(x+3)\cdot(x-3)}+x-3}\\\\\mathsf{2x-2x-8=2\cdot\sqrt{x^2-3x+3x-9}}$

$\mathsf{\dfrac{-8}{2}=\sqrt{x^2-9}}\\\\\mathsf{(-4)^2=(\sqrt{x^2-9})^2}\\\\\mathsf{16=x^2-9}\\\\\mathsf{x^2-9=16~\gets~invertido}\\\\\mathsf{x^2=25}\\\\\mathsf{\sqrt{x^2}=\sqrt{25}}\\\\\mathsf{x=\pm5}$


Pela condição de existência, x deve ser maior que 4, portanto a solução dessa equação é:


$\boxed{\boxed{\mathsf{x=5}}}$


Dúvidas? comente.