Hi Emre,
a) 2log 32 = log2 2^5 = 5*log2(2) = 5
b) 2log 64 = 6
c) 10log 1000000 = log10(10^6) = 6
d) 2log 1 = 0 (loga(1) = 0 gilt immer ;) )
a) log (abcd) = log(a)+log(b)+log(c)+log(d)
b) log (abc /d) = log(a)+log(b)+log(c)-log(d)
c) log (ad/bc) = log(a)+log(d)-log(b)-log(c)
d) log (x2yz3) = log(x^2)+log(y)+log(z^3) = 2log(x)+log(y)+3log(z)
Das sind eigentlich immer direkt die Logarithmengesetze verwendet ;).
Grüße