Answer:
a)
a1 = log(1) = 0 (2⁰ = 1)
a2 = log(2) = 1 (2¹ = 2)
a3 = log(3) = ln(3)/ln(2) = 1.098/0.693 = 1.5849
a4 = log(4) = 2 (2² = 4)
a5 = log(5) = ln(5)/ln(2) = 1.610/0.693 = 2.322
a6 = log(6) = log(3*2) = log(3)+log(2) = 1.5849+1 = 2.5849 (here I use the property log(a*b) = log(a)+log(b)
a7 = log(7) = ln(7)/ln(2) = 1.9459/0.6932 = 2.807
a8 = log(8) = 3 (2³ = 8)
a9 = log(9) = log(3²) = 2*log(3) = 2*1.5849 = 3.1699 (I use the property log(a^k) = k*log(a) )
a10 = log(10) = log(2*5) = log(2)+log(5) = 1+ 2.322= 3.322
b) I can take the results of log n we previously computed above to calculate 2^log(n), however the idea of this exercise is to learn about the definition of log_2:
log(x) is the number L such that 2^L = x. Therefore 2^log(n) = n if we take the log in base 2. This means that
a1 = 1
a2 = 2
a3 = 3
a4 = 4
a5 = 5
a6 = 6
a7 = 7
a8 = 8
a9 = 9
a10 = 10
I hope this works for you!!
Answer:
-4,0 -4,0
Step-by-step explanation:
-4,0 -4,0
Answer:
17e+18
Step-by-step explanation:
9e+4=-5e+14+13e
4e+4+14+13e
17e+4+14
17e+18
A has fixed one time fee of $12 and if you go to it say "m" months you pay $28 for each month, so your total cost at A is really 12 + 28m.
B has a fixed one time fee of $20 and if you go to it "m" months you pay $26 for each month, so you total cost at B is 20 + 26m.
how many months for the cost to be the same?
well, since the cost for both is the same, we can just get A's, knowing that B is the same
