I.
1) log3 243 = x
2) log1/2 32 = x
3) log1/6 216 = x
4) log2 0,125 = x
5) log 0,001 = x
6)log1/3 1 = x
Bo każda liczba podniesiona do potęgi zerowej da nam 1
II.
1) log 25 + log 4 = log (25*4) = log 100 = 2
2) log 5000 - log 5 = log (5000/5) = log 1000 = 3
3) 4log 2 + 2log 25 =
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I.
1) log3 243 = x
2) log1/2 32 = x
3) log1/6 216 = x
4) log2 0,125 = x
5) log 0,001 = x
6)log1/3 1 = x
Bo każda liczba podniesiona do potęgi zerowej da nam 1
II.
1) log 25 + log 4 = log (25*4) = log 100 = 2
2) log 5000 - log 5 = log (5000/5) = log 1000 = 3
3) 4log 2 + 2log 25 =