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Question 117028: suppose that the function P=12+50ln x represents the percentage of inbound e-mail in the U.S> that is considered spam, where x is the number of years after 2001.
carry all calculationjs to six decimals on each intermediate step, when necessary.
use this model to approximate the percentage of spam in the year 2005
Answer by bucky(2189)   (Show Source):
You can put this solution on YOUR website! Given the equation:
.
.
In which P represents the percent (as a whole number) of the inbound e-mail in the U.S. that is spam,
and x represents the number of years after 2001 that will determine what P is.
.
For this problem you are asked to solve for x if P is 90 percent. Begin by substituting
90 for P in the equation to get:
.
.
Get rid of the 12 on the right side by subtracting 12 from both sides. When you do that subtraction,
you reduce the equation to:
.
.
Just to get this into a little more standard form ... with the unknown on the left side ...
transpose (switch sides) this equation to:
.
.
Divide both sides by 50 to reduce the equation to:
.
.
which (leaving out the the step that shows 78 divided by 50} is just:
.
.
This is a logarithmic form of an equation. It can be converted to an equivalent exponential form
by applying a rule that says the logarithmic form:
.
.
is equivalent to the exponential form 
.
In this problem b, the base of the logarithm, is e which is the base of the natural logarithms
that are commonly indicated by "ln". By comparing the logarithmic form of this rule to the logarithmic
form derived in our problem, you can see that b = e, y = 1.56, and x = x. Substitute these
values into the exponential form of the rule and you get:
.
.
as being the exponential form that is equivalent to
.
You can use the exponential form to solve for x. Transpose the exponential form to get the
x in the more standard arrangement of having the unknown on the left side and the exponential
form becomes:
.
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You can now use a scientific calculator to solve for x. Just think of e as being a number that
you are going to raise to the 1.56 power. [The value of e is approximately 2.718281828]
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So you can use your calculator to raise 2.718281828 to the 1.56 power. Or look on your calculator
for the function key that shows  . If you have one, your calculator might work
this way: (1) enter 1.56, then (2) activate the key.
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Either way you should get an answer of 4.758821245 which rounds off to 4.76. This tells you
that 4.76 years after 2001 the inbound spam in the U.S. can be expected to be 90 percent of
the message traffic.
.
(Assuming that this equation begins on December 31, 2001 ... 4.76 years later
(4 and three-quarters years later) the spam level would reach 90 percent around September 30, 2006
which is nearly 5 years after December 31, 2001 ... actually 4.75 years after December 31, 2001
...
.
Hope this helps you to see your way through this problem and helps you also to learn a little
about working with logarithms ... especially natural logarithms (indicated by ln).
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