Question

Quais são os valores de $\mathsf{x}$ que satisfazem a desigualdade abaixo:

$\mathsf{\log_6 (x - 3) \ \textless \ \log_7 (x - 3)}$


a) $\mathsf{x \ \textgreater \ \frac{6}{7}}$

b) $\mathsf{x \ \textless \ 1}$

c) $\mathsf{3 \ \textless \ x \ \textless \ 4}$

d) $\mathsf{x \ \textless \ 3}$

e) $\mathsf{x \ \textless \ 4}$

Answer 1

$\large\begin{array}{l} \textsf{Podemos usar a lei de mudan\c{c}a de base:}\\\\ \mathsf{\ell og_b\,a=\dfrac{\ell og_c\,a}{\ell og_c\,b}} \end{array}$


$\large\begin{array}{l} \textsf{Vamos mudar os logaritmos envolvidos na inequa\c{c}\~ao para}\\\textsf{uma base arbitr\'aria. Por exemplo, base 10:}\\\\ \mathsf{\ell og_6\,(x-3)<\ell og_7\,(x-3)}\\\\ \mathsf{\dfrac{\ell og(x-3)}{\ell og\,6}<\dfrac{\ell og(x-3)}{\ell og\,7}}\\\\ \mathsf{\dfrac{\ell og(x-3)}{\ell og\,6}-\dfrac{\ell og(x-3)}{\ell og\,7}<0}\\\\ \mathsf{\ell og(x-3)\cdot \left(\dfrac{1}{\ell og\,6}-\dfrac{1}{\ell og\,7}\right)<0} \end{array}$

$\large\begin{array}{l} \mathsf{\ell og(x-3)\cdot \left(\dfrac{\ell og\,7}{\ell og\,6\cdot \ell og\,7}-\dfrac{\ell og\,6}{\ell og\,6\cdot \ell og\,7}\right)<0}\\\\ \mathsf{\ell og(x-3)\cdot\dfrac{\ell og\,7-\ell og\,6}{\ell og\,6\cdot \ell og\,7}<0\qquad(i)} \end{array}$


$\large\begin{array}{l} \textsf{Sabemos que}\\\\ \mathsf{1<6<7}\\\\ \textsf{e como a base 10 \'e maior que 1,}\\\\ \mathsf{\ell og\,1<\ell og\,6<\ell og\,7}\\\\ \mathsf{0<\ell og\,6<\ell og\,7\qquad(ii)}\\\\ \textsf{(os logaritmos de 6 e 7 s\~ao positivos na base 10)} \end{array}$


$\large\begin{array}{l} \textsf{Ainda por (ii),}\\\\ \mathsf{\ell og\,6<\ell og\,7}\\\\ \mathsf{\ell og\,7-\ell og\,6>0\qquad(iii)}\\\\\\ \textsf{e tamb\'em}\\\\ \mathsf{\ell og\,6\cdot \ell og\,7>0\qquad(iv)} \end{array}$


$\large\begin{array}{l} \textsf{Por (iii) e (iv), conclui-se que}\\\\ \mathsf{\dfrac{\ell og\,6\cdot \ell og\,7}{\ell og\,7-\ell og\,6}>0\qquad(v)} \end{array}$


$\large\begin{array}{l} \textsf{Voltando \`a inequa\c{c}\~ao (i), multiplicando ambos os lados por}\\\\ \mathsf{\dfrac{\ell og\,6\cdot \ell og\,7}{\ell og\,7-\ell og\,6}}\\\\ \textsf{que \'e positivo, o sentido da desigualdade \'e preservado:} \end{array}$

$\large\begin{array}{l} \mathsf{\ell og(x-3)\cdot\dfrac{\ell og\,7-\ell og\,6}{\ell og\,6\cdot \ell og\,7}\cdot \dfrac{\ell og\,6\cdot \ell og\,7}{\ell og\,7-\ell og\,6}<0\cdot \dfrac{\ell og\,6\cdot \ell og\,7}{\ell og\,7-\ell og\,6}}\\\\ \mathsf{\ell og(x-3)\cdot 1<0}\\\\ \mathsf{\ell og(x-3)<0} \end{array}$


$\large\begin{array}{l} \textsf{Como a base do logaritmo \'e }\mathsf{10>1,}\textsf{ a fun\c{c}\~ao logar\'itmica}\\\\ \mathsf{f(x)=\ell og(x-3)}\\\\ \textsf{\'e crescente. Se}\\\\ \mathsf{\ell og(x-3)<0}\\\\ \textsf{ent\~ao}\\\\ \mathsf{0<x-3<10^0}\\\\ \mathsf{0<x-3<1}\\\\\mathsf{0+3<x<1+3} \end{array}$

$\large\begin{array}{l} \boxed{\begin{array}{c}\mathsf{3<x<4} \end{array}} \end{array}$   <———    $\large\textsf{esta \'e a resposta (alternativa c).}$


$\large\begin{array}{l} \textsf{D\'uvidas? Comente.}\\\\\\ \textsf{Bons estudos! :-)} \end{array}$