Question 117420
If you started with {{{y[0]}}} amount of
pollution, and reduced it by 50%, you'd have
{{{.5*y[0]}}} left, so
{{{y = .5*y[0]}}}
{{{y = y[0]*e^(-.3821*t)}}}
{{{.5*y[0] = y[0]*e^(-.3821*t)}}}
divide both sides by {{{y[0]}}}
{{{.5 = e^(-.3821*t)}}}
Find the ln (log to the base e) of both sides
{{{ln(.5) = ln(e^(-.3821*t))}}}
The trick is to READ the right side correctly. What does it say,exactly?
It says "What log to the base e gives me {{{e^(-.3821*t)}}}?
It's very obviously {{{-.3821*t}}}, so
{{{ln(.5) = -.3821*t}}}
Use a calculator to find the left side
{{{-.69315 = -.3821*t}}}
{{{t = 1.814}}}years
{{{.814}}}years X {{{12}}} months/year = {{{9.768}}}
So far its
1 year 9.768 months
{{{.768}}}months X {{{30}}}days/month = {{{23.04}}}
So, the answer is
1 year, 9 months, 23 days approximately 
check answer
{{{.5 = e^(-.3821*t)}}}
{{{.5 = e^(-.3821*1.814)}}}
{{{.5 = e^-.6931}}}
{{{.5 = .50000889}}} this is close enough