Question 92738
the function P = 13 + 45*ln(x) represents the percentage of inbound e-mail in the U.S. that is considered spam, where x is the number of years after 2000.
Carry all calculations to six decimals on each intermediate step when necessary.
:
a) Use this model to approximate the percentage of spam in the year 2003.
P = 13 + 45*ln(x)
Substitute 3 for x (for 2003) and find P
P = 13 + 45*ln(3)
P = 13 + 45 * 1.098612; (the nat log of 3)
P = 13 + 49.437553
P = 62.437553 % is spam in 2003
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b) Use this model to approximate the year that the percent of spam will reach 95% provided that law enforcement regarding spammers does not change.
:
Substitute 95 for P, and find x:
13 + 45*ln(x) = 95
45*ln(x) = 95 - 13; subtract 13 from both sides
45*ln(x) = 82
ln(x) = {{{82/45)}}}; divide both sides by 45
x = 6.18559; find e^(82/45) on a calc
:
in 2006, 95% will be spam
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How about this, did it make you a little wiser? Any questions about this procedure?