Does anyone get this?

Question

5)Suppose that the function P = 13+45ln 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.

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.

Answer 1

Yes. many people do.

a)

P = 13 + 45 * ln (x) = 13 + 45 * ln (2003 - 2000)
= 13 + 45 * ln(3) = 62.4376

b)

let y = year spam is 95%

95 = 13 + 45 * ln (y - 2000)

82 / 45 = ln (y-2000)

exp (82/45) = exp (ln (y-2000)) = y -2000

exp (82/45) + 2000 = y
y = 6.18559 + 2000 ≈ 2006

Answer 2

P = 13 + 45*ln(x)

a) 2003 is year 3. P = 13 + 45*ln(3), which is a computation you can do with your calculator.

b) This one is an algebra problem.

95 = 13 + 45*ln(x)
82 = 45*ln(x)
82 / 45 = ln(x)
e^(82 / 45) = x

This is another computation for your calculator. Then add x to 2000 and round to the nearest year.

Answer 3

a) p=13+45ln(3) -solve
p=62.437553
62.437553%

b) 95=13+45ln x -subtract 13 from both sides
82=45ln x -divide both sides by 45
1.822222=ln x -put each side as a power of e.
6.185588=x
Year 2006