zad 9[tex]3 log_5 2 + log_2 7 = log_5 2^3 + log_2 7 = log_5 (8 \times 7) = log_5 56[/tex]
zad 10jesli to mialo byc [tex]2^x[/tex] to dodajemy wektor [-3,0] i mamy[tex]2^{x+3} = 2^x \times 2^3 = 2^x \times 8[/tex]
w obecnej postaci zad 10 wykracza poza material z liceum
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zad 9
[tex]3 log_5 2 + log_2 7 = log_5 2^3 + log_2 7 = log_5 (8 \times 7) = log_5 56[/tex]
zad 10
jesli to mialo byc [tex]2^x[/tex] to dodajemy wektor [-3,0] i mamy
[tex]2^{x+3} = 2^x \times 2^3 = 2^x \times 8[/tex]
w obecnej postaci zad 10 wykracza poza material z liceum