LOGARYTMY     Proszę o rozwiązanie i wytłumaczenie sposobu robienia tych przykładów. Zadanie w załączniku.

a) [latex]log_{14} 2 = a i log_{14} 5 = b[/latex] [latex]log_7 50 =frac{log_{14} 50}{log_{14} 7} =frac{log_{14} (2 cdot 25)}{log_{14} frac{14}{2}} =frac{log_{14} 2 + log_{14} 25}{log_{14} 14 - log_{14} 2} =frac{a+ log_{14} 5^2}{1 - a} = \\ =frac{a+ 2 cdot log_{14} 5}{1 - a} =frac{a+ 2b}{1 - a}[/latex] [latex]log_7 50 =frac{a+ 2b}{1 - a}[/latex] b) [latex]log_3 20 = a i log_3 15 = b[/latex] [latex]log_3 15 = b \ log_3 15 = log_3 (3 cdot 5) = log_3 3 + log_3 5 =1 +log_3 5 \\ 1 +log_3 5 = b \ log_3 5 = b - 1[/latex] [latex]log_3 20 = a \ log_3 20 = log_3 (4 cdot 5) = log_3 4 + log_3 5 =log_3 2^2 +log_3 5 = \ = 2 dot log_3 2 +log_3 5 \\ 2 cdot log_3 2 +log_3 5 = a \ 2 cdot log_3 2 = a - log_3 5 \ 2 cdot log_3 2 = a - (b-1) \ 2 cdot log_3 2 = a - b+1 /:2 \ log_3 2 =frac{a - b+1}{2}[/latex] [latex]log_2 360 = frac{log_3 360}{log_3 2}= frac{log_3 (2 cdot 9 cdot 20)}{frac{a - b+1}{2}} = frac{log_3 2 + log_3 9 + log_3 20}{frac{a - b+1}{2}}= \\ = frac{frac{a - b+1}{2}+ log_3 3^2 + a}{frac{a - b+1}{2}}= frac{frac{a - b+1}{2}+ 2 cdot log_3 3 + frac{2a}{2}}{frac{a - b+1}{2}}= frac{frac{a - b+1}{2}+ 2 cdot 1 + frac{2a}{2}}{frac{a - b+1}{2}}=[/latex] [latex]= frac{frac{a - b+1}{2}+ 2 + frac{2a}{2}}{frac{a - b+1}{2}}= frac{frac{a - b+1}{2}+ frac{4}{2} + frac{2a}{2}}{frac{a - b+1}{2}}=frac{frac{a - b+1+4+2a}{2}}{frac{a - b+1}{2}}=frac{frac{3a - b+5}{2}}{frac{a - b+1}{2}}= \\ =frac{3a - b+5}{2} :frac{a - b+1}{2}=frac{3a - b+5}{2} cdot frac{2}{a - b+1}=frac{3a - b+5}{a - b+1}[/latex] [latex]log_2 360 =frac{3a - b+5}{a - b+1}[/latex]