Question

lg2/1+lg3/2+...............+lg10/9=?

Answer 1

$\lg\Big(\dfrac{2}{1}\Big) + \lg\Big(\dfrac{3}{2}\Big) + ...+\lg\Big(\dfrac{10}{9}\Big) = \\ \\ =\Big(\lg2-\lg1\Big)+ \Big(\lg3-\lg2\Big)+ ...+\Big(\lg10-\lg9\Big)= \\ \\ = \lg2+\lg3+...+\lg10 - \lg1-\lg2-\lg3-...-\lg10 = \\ \\ = \lg2+\lg3+...+\lg10 -(\lg1 +\lg2+\lg3+...+\lg10) = \\ \\ = \lg2+\lg3+...+\lg10 -(0 +\lg2+\lg3+...+\lg10) = \\ \\ = \lg2+\lg3+...+\lg10 -(\lg2+\lg3+...+\lg10) = \\ \\ = 0$


$\\ M-am folosit de proprietatea: \log_{\big a}b-\log_{\big a}c = \log_{\big a} \Big(\dfrac{b}{c}\Big) \\ \\ La noi baza era 10:\quad \lg a = \log_{\big{10}}a$

Answer 2

Lg2/1+lg3/2+...............+lg10/9=?

$\it lg\dfrac{2}{1}+lg\dfrac{3}{2}+lg\dfrac{4}{3}+ ... lg\dfrac{10}{9}= lg\left(\dfrac{2}{1}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\ ...\ \cdot\dfrac{10}{9}\right) =lg10=1$