Question

Dado (a - b ≠ 1) e (a + b ≠ 1), mostre que 1 / log[a+b] (c) + 1 / log[a-b] (c) = 2

Answer 1

$\frac{1}{log_{(a+b)}c} +\frac{1}{log_{(a-b)}c} =2 \Rightarrow \\ \\ Propriedade{:} \\ log_n(m)= \frac{log(m)}{log(n)} \\ \\ \\ log_{(a+b)}c= \frac{logc}{log(a+b)} \\ \\ log_{(a-b)}c= \frac{logc}{log(a-b)} \\ \\ \\ \frac{1}{\frac{logc}{log(a+b)}} +\frac{1}{\frac{logc}{log(a-b)}} =2\Rightarrow \\ \\ \\ \frac{log(a+b)}{logc}} +\frac{log(a-b)}{logc} =2 \\ \\ log(a+b)+log(a-b)=2.logc \\ \\ Propriedades{:} \\ log(m*n)=log(m)+log(n) \\ log(m)^n=n.log(m) \\ \\ log[(a+b)(a-b)]=logc^2$
$Propriedade{:} \\ Quadrado ~da~diferen\c{c}a{:} \\ (m+n)(m-n)=(m^2-n^2) \\ \\ \\ (a+b)(a-b)=c^2 \\a^2-b^2=c^2 ==\ \textgreater \ a^2=b^2+c^2 ~(teorema~ de~ Pita\´goras)$

Answer 2

Olá!!

$\\ \displaystyle \mathsf{\frac{1}{\log_{a + b} c} + \frac{1}{\log_{a - b} c} =} \\\\ \Downarrow \\\\ \mathsf{\frac{1}{\frac{\log c}{\log (a + b)}} + \frac{1}{\frac{\log c}{\log (a - b)}} =} \\\\ \Downarrow \\\\ \mathsf{1 \div \frac{\log c}{\log (a + b)} + 1 \div \frac{\log c}{\log (a - b)} =}$

$\\ \displaystyle \Downarrow \\\\ \mathsf{1 \cdot \frac{\log (a + b)}{\log c} + 1 \cdot \frac{\log (a - b)}{\log c} =} \\\\ \Downarrow \\\\ \mathsf{\frac{\log (a + b) + \log (a - b)}{\log c} =} \\\\ \Downarrow \\\\ \mathsf{\frac{\log \left [ (a + b) \cdot (a - b) \right ]}{\log c} =}$

$\\ \Downarrow \\\\ \mathsf{\frac{\log (a^2 - b^2)}{\log c} =}$

 Aplicando o Teorema de Pitágoras no triângulo em questão, tiramos que:

$\\ \mathrm{a^2 = b^2 + c^2} \\\\ \mathrm{a^2 - b^2 = c^2}$

 Portanto, substituindo...

$\\ \displaystyle \mathsf{\frac{\log (a^2 - b^2)}{\log c} =} \\\\ \Downarrow \\\\ \mathsf{\frac{\log c^2}{\log c} =} \\\\ \Downarrow \\\\ \mathsf{\frac{2 \cdot \log c}{\log c} =} \\\\ \Downarrow \\\\ \boxed{\mathsf{2}}$

 Que é o que queríamos demonstrar!!


Propriedades de logaritmos utilizadas:

$\\ \bullet \qquad \mathbf{\log_b a = \frac{\log a}{\log b};} \\\\ \bullet \qquad \mathbf{\log_b (a \cdot c) = \log_b a + \log_b c;} \\\\ \bullet \qquad \mathbf{\log_b a^n = n \cdot \log_b a.}$