type 1 log(xy) = log x + log y
type 2 log (x)^y = y log x
now look at a) type 1, true
b) not type 1, not type 2, false
in fact log 3x = log 3 + log x
c) type 1 and type 2
log (3*2^2* y) = log 3 + 2 log 2 + log y = log 3y + 2 log 2 true
d) type 1
log(20y/4) = log(20y) - log(4)
type 3 base^logbase(x) = x
g) ln(1) = log(1)
e^ln 1 = 1 = 10^log 1
ln 1 = 0 = log 1 true type 3
h) ln(e) = log(10) ?????????
e^lne = e so lne =1
also log 10=1
so true, type 3
i) y = e^-3 do not forget the ^ for exponent
ln y = -3 ln e
ln y = -3 not what you want
j) 10^log x = 10^-12
x = 10^-12 true but remember ^ sign for exponents !
k) e^ln x = x = e^-12 not 10^-12 false
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