Question 404254
{{{P = 12 +20* ln(x)}}}
First, you have to find out what {{{P}}} is in the year 2001
If {{{x}}} is the number of years after 2000, then {{{x = 1}}} when
the year = 2001, so
{{{P = 12 + 20* ln(1)}}}
{{{P = 12 + 20* 0}}}
{{{P = 12}}} (or, 12%)
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I want to get an idea how fast {{{P}}} increases with {{{x}}}
{{{x = 2}}}, {{{P = 12 + 20*.693}}} (25.863% in 2002)
{{{x = 3}}}, {{{P = 12 + 20*1.099}}} (33.972% in 2003)
{{{x = 4}}}, {{{P = 12 + 20*1.386}}} (39.726% in 2004)
{{{x = 5}}}, {{{P = 12 + 20*1.609}}} (44.189% in 2005)
OK, {{{x}}} is going to be pretty large.
{{{90 = 12 + 20*ln(x)}}}
{{{20*ln(x) = 78}}}
{{{ln(x) = 78/20}}}
{{{ln(x) = 3.9}}}
I'll write this as an exponential:
{{{x = e^3.9}}}
{{{x = 49.40}}}
So, it takes 49.4 years after 2000, or 2049 + 5 months to reach 90% spam
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Now find {{{P}}} when {{{x = 5}}}
{{{P = 12 + 20*ln(5)}}}
{{{P = 12 + 20*1.609}}}
{{{P = 12 + 32.189}}}
{{{P = 44.19}}}
The spam is 44.19% in 2005
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I would guess that your problem is:
When they say percent, they don't mean .90, but the actual %, 90