Zamiana podstawy logarytmu.
: 27 lis 2017, 19:36
3 f)
\(p=log_32\\a=log_{ \sqrt[3]{3} }6= \frac{log_36}{log_33^{ \frac{1}{3} }} = \frac{log_3(2 \cdot 3)}{ \frac{1}{3} }= \frac{log_32+log_33}{ \frac{1}{3} }= \frac{p+1}{ \frac{1}{3} }=3p+3\)
5 f)
\(a=log_23\\
\frac{1}{log_{ \sqrt{3} }2} + \frac{1}{log_94}= \frac{1}{\frac{log_22}{log_23^{ \frac{1}{2}} }}+ \frac{1}{ \frac{log_22^2}{log_29} }=\frac{1}{ \frac{1}{ \frac{1}{2}log_23 } }+ \frac{1}{ \frac{2}{log_23^2} }= \frac{1}{ \frac{1}{\frac{a}{2}} }+ \frac{1}{ \frac{2}{2a} }= \frac{a}{2}+a=1,5a\)
\(p=log_32\\a=log_{ \sqrt[3]{3} }6= \frac{log_36}{log_33^{ \frac{1}{3} }} = \frac{log_3(2 \cdot 3)}{ \frac{1}{3} }= \frac{log_32+log_33}{ \frac{1}{3} }= \frac{p+1}{ \frac{1}{3} }=3p+3\)
5 f)
\(a=log_23\\
\frac{1}{log_{ \sqrt{3} }2} + \frac{1}{log_94}= \frac{1}{\frac{log_22}{log_23^{ \frac{1}{2}} }}+ \frac{1}{ \frac{log_22^2}{log_29} }=\frac{1}{ \frac{1}{ \frac{1}{2}log_23 } }+ \frac{1}{ \frac{2}{log_23^2} }= \frac{1}{ \frac{1}{\frac{a}{2}} }+ \frac{1}{ \frac{2}{2a} }= \frac{a}{2}+a=1,5a\)