Question

Proszę o pomoc - logarytmy

Answer 1

Odpowiedź:

a )  = [tex]log 5^2 - log 10 = log 25 - log 10 = log \frac{25}{10} = log 2,5[/tex]

b)  = 0,5*( [tex]log_3 3 + log_3 2^3) = 0,5 *( log_3 3 + log_3 8 ) = 0,5*log_3 (3*8) = 0,5 log_3 24[/tex] = [tex]log_3 24^{0,5} = log_3 \sqrt{24} c )[/tex][tex]\\\\\\log_5 27^{2/3} + log_5 32^{2/5} = log_5 9 + log_5 4 = log_5 (9*4) = log_5 36[/tex]

d )  = [tex]log_3 ( 2^{1/3})^{12/5} - log_3 ( 4^{1/3})^{3/2} = log_3 2^{4/5}- log_3 4^{1/2} =[/tex]

=  [tex]log_3 ( 2^{4/5} : 2^1 ) = log_3 2^{-1/5}[/tex]

e)  = [tex]\frac{log_2 4 + log_2 27}{3} = \frac{log_2 ( 4*27)}{3} = \frac{1}{3} log_2 108 =[/tex] [tex]log_2 108^{1/3}[/tex] [tex]= log_2 (3*\sqrt[3]{4})[/tex]

f )  = [tex]\frac{-2 log_4 4 + log_4 36^{1/2}}{log_4 4 - log_4 0,25} =[/tex] [tex]\frac{log_4 4^{-2} + log_4 6}{log_4 ( 4 : 0,25)} = \frac{log_4 \frac{6}{16} }{log_4 16} =[/tex]

Szczegółowe wyjaśnienie:[tex]\frac{log_4 \frac{3}{8} }{2} = 0,5 log_4 \frac{3}{8} = log_4 ( \frac{3}{8})^{0,5} =[/tex] log_4 [tex](\frac{\sqrt{3} }{2\sqrt{2} }) = log_4 ( \frac{\sqrt{6} }{4})[/tex]