Question

Sabendo que as coordenadas X e Y são (x,0) calcule X

F(x)=㏒ₓ4+㏒₄x-5/2

Answer 1

Olá Spawwn.


Propriedades logarítmicas usadas:


$\star~~\boxed{\boxed{\mathsf{\ell og_ba=c\Rightarrow b^c=a}}}\\\\\\\\\star~~\boxed{\boxed{\mathsf{\ell og_ab\Rightarrow\dfrac{\ell og_cb}{\ell og_ca}}}}$


Organizando e resolvendo a equação:


$\mathsf{f(x)=y=\ell og_x4+\ell og_4x-\dfrac{5}{2}}\\\\\\\mathsf{0=\ell og_x4+\ell og_4x-\dfrac{5}{2}~~~\gets~~~~(x,0)}\\\\\\\underline{\qquad\qquad\qquad\qquad\qquad}\\\\\\\mathsf{\star~~\ell og_x4=\dfrac{\ell og_44}{\ell og_4x}\Rightarrow \dfrac{1}{\ell og_4x}}\\\\\\\mathsf{\star~~\ell og_4a=\dfrac{5}{2}~\Rightarrow 4^{\frac{5}{2}}=a~\Rightarrow (2^2)^\frac{5}{2}=a~\Rightarrow 2^5=a~\Rightarrow 32 = a}\\\\\\\underline{\qquad\qquad\qquad\qquad\qquad}$


$\mathsf{0=\dfrac{1}{\ell og_4x}+\ell og_4x-\ell og_432~\cdot(\ell og_4x)}\\\\\\\mathsf{0=1+(\ell og_4x)^2-\ell og_432\cdot \ell og_4x}\\\\\mathsf{(\ell og_4x)^2-\ell og_432\cdot \ell og_4x+\ell og_44=0~~\gets~~organizando.}\\\\\\\\\mathsf{\Delta=(\ell og_432)^2-4\cdot 1\cdot\ell og_44}\\\\\mathsf{\Delta=\Big(\dfrac{5}{2}\Big)^2-4}\\\\\mathsf{\Delta=\dfrac{25-16}{4}}\\\\\mathsf{\Delta=\dfrac{9}{4}}\\\\\mathsf{\Delta=\Big(\dfrac{3}{2}\Big)^2}$


$\mathsf{\ell og_4x=\dfrac{-\Big(-\dfrac{5}{2}\Big)\pm\sqrt{\Big(\dfrac{3}{2}\Big)^2}}{2\cdot 1}}\\\\\\\mathsf{\ell og_4x^+=\dfrac{\dfrac{5}{2}+\dfrac{3}{2}}{2}\qquad\qquad\qquad\qquad \ell og_4x^-=\dfrac{\dfrac{5}{2}-\dfrac{3}{2}}{2}}\\\\\\\\\mathsf{\ell og_4x^+=\dfrac{\diagdown\!\!\!\!8}{\diagdown\!\!\!\!2}\cdot\dfrac{1}{2}\qquad\qquad\qquad\qquad~~\ell og_4x^-=\dfrac{\diagup\!\!\!\!2}{\diagup\!\!\!\!2}\cdot\dfrac{1}{2}}\\\\\\\\\mathsf{\ell og_4x^+=2\qquad\qquad\qquad\qquad\qquad\ell og_4x^-=\dfrac{1}{2}}$


$\mathsf{\ell og_4x=2\Rightarrow 4^2=x\Rightarrow \boxed{\mathsf{16 = x}}}\\\\\\\mathsf{\ell og_4x=\dfrac{1}{2}\Rightarrow 4^{\frac{1}{2}}=x\Rightarrow \boxed{\mathsf{2=x}}}$


Solução do par ordenado é (2, 0) ou (16, 0).


Dúvidas? comente.