Strona 1 z 1
LOGARYTMY
: 11 mar 2017, 16:10
autor: mtworek98
Oblicz:
a) \(9\log_{9}2 - \log0,2 + \log25\)
b) \(4^{2\log_{2}3} + \log_{2}5\)
c) \(\frac{2\log_{3}4 - 3\log_{3}12}{3 + \log_{3}4}\)
Re: LOGARYTMY
: 12 mar 2017, 10:34
autor: eresh
mtworek98 pisze:Oblicz:
b) \(4^{2\log_{2}3} + \log_{2}5\)
\(4^{2\log_23}+\log_25=4^{\log_23^2}+\log_25=2^{2\log_29}+\log_25=2^{\log_281}+\log_25=81+\log_25\)
Re: LOGARYTMY
: 12 mar 2017, 10:35
autor: eresh
mtworek98 pisze:Oblicz:
a) \(9\log_{9}2 - \log0,2 + \log25\)
\(9\log_92-\log 0,2+\log 25=9\log_92+\log\frac{25}{0,2}=9\log_92+\log 125=9\log_92+3log 5=3(3\log_92+\log 5)\)
Re: LOGARYTMY
: 12 mar 2017, 10:41
autor: eresh
mtworek98 pisze:Oblicz:
c) \(\frac{2\log_{3}4 - 3\log_{3}12}{3 + \log_{3}4}\)
\(\frac{2\log_34-3\log_312}{3+\log_34}=\frac{2\log_34-3\log_3(3\cdot 4)}{3+\log_34}=\frac{2\log_34-3-3\log_34}{3+\log_34}=-\frac{\log_34+3}{\log_34+3}=-1\)