Question

Logaritmi

Rezolvare pas cu pas ​

Answer 1

Răspuns:

Explicație pas cu pas:

.

Spor

Answer 2

$\it b^2+c^2=a^2 \Rightarrow c^2=a^2-b^2\ \ \ \ \ (*)\\ \\ \\ log_{(a+b)} c+log_{(a-b)} c=2log_{(a+b)} c\cdot log_{(a-b)} c\ \Leftrightarrow\\ \\ \\ \Leftrightarrow \dfrac{log_c c}{log_c (a+b)} +\dfrac{log_c c}{log_c (a-b)}=2\dfrac{log_c c}{log_c (a+b)}\cdot\dfrac{log_c c}{log_c (a-b)} \Leftrightarrow\\ \\ \\ \Leftrightarrow \dfrac{1}{log_c (a+b)} +\dfrac{1}{log_c (a-b)}=2\dfrac{1}{log_c (a+b)}\cdot\dfrac{1}{log_c (a-b)} \Leftrightarrow$

$\it \Leftrightarrow \dfrac{log_c (a-b)+log_c(a+b)}{log_c (a+b) \cdot log_c(a-b)}=2\dfrac{1}{log_c (a+b)\cdot log_c(a-b)} \Leftrightarrow \\ \\ \\ \Leftrightarrow \dfrac{log_c (a-b)(a+b)}{log_c (a+b) \cdot log_c(a-b)}=\dfrac{2}{log_c (a+b)\cdot log_c(a-b)} \Leftrightarrow\\ \\ \\ \Leftrightarrow \dfrac{log_c (a^2-b^2)}{log_c (a+b) \cdot log_c(a-b)}=\dfrac{2}{log_c (a+b)\cdot log_c(a-b)} \stackrel{(*)}{\Longleftrightarrow}$

$\it \Leftrightarrow \dfrac{log_c c^2}{log_c (a+b) \cdot log_c(a-b)}=\dfrac{2}{log_c (a+b)\cdot log_c(a-b)} \Leftrightarrow\\ \\ \\ \Leftrightarrow \dfrac{2}{log_c (a+b)\cdot log_c(a-b)} =\dfrac{2}{log_c (a+b)\cdot log_c(a-b)} \ \ \ (A)$