Czy x jest liczbą całkowitą? Zapisz obliczenia!!!!
Korzystam z własności logarymów
log(a*b) = loga +logb
log_a z a = 1,a dokładniej log10 = 1
x = log2* log(5*10) + (log5)^2 = log2*(log5 +log10) +(log5)^2= log2*(log5 +1) +(log5)^2 =
= log2*log5 +log2 + (log5)^2 = log5(log2 + log5) + log2 = log5*log(2*5) + log2
= log5*log10 + log2 = log5*1 + log 2 = log5 + log2 = log(5*2) = log10 = 1
x = 1, czyli x jest liczbą całkowitą
Myślę, że pomogłam :-)
log2*log50 + log5*log5 = log2*log(5*10) + log5*log5 = log2*(log5 + log10) + log5*log5 =
= log2*(log5 + 1) + log5*log5 = log2*log5 + log2 + log5*log5 =
= log5*(log2 + log5) + log2 = log5*log(2*5) + log2 = log5*log10 + log2 =
= log5 + log2 = log(2*5) = log10
Uwaga
W zadaniu wykorzystano
log(a*b) = loga + logb
oraz
log10 = 1
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Korzystam z własności logarymów
log(a*b) = loga +logb
log_a z a = 1,a dokładniej log10 = 1
x = log2* log(5*10) + (log5)^2 = log2*(log5 +log10) +(log5)^2= log2*(log5 +1) +(log5)^2 =
= log2*log5 +log2 + (log5)^2 = log5(log2 + log5) + log2 = log5*log(2*5) + log2
= log5*log10 + log2 = log5*1 + log 2 = log5 + log2 = log(5*2) = log10 = 1
x = 1, czyli x jest liczbą całkowitą
Myślę, że pomogłam :-)
log2*log50 + log5*log5 = log2*log(5*10) + log5*log5 = log2*(log5 + log10) + log5*log5 =
= log2*(log5 + 1) + log5*log5 = log2*log5 + log2 + log5*log5 =
= log5*(log2 + log5) + log2 = log5*log(2*5) + log2 = log5*log10 + log2 =
= log5 + log2 = log(2*5) = log10
Uwaga
W zadaniu wykorzystano
log(a*b) = loga + logb
oraz
log10 = 1