document.write( "Question 1154016: A small lake is stocked with a certain species of fish. The fish population is modelled by the function P(t) = 10/1+5e^(-0.8t) where P is the number of fish in hundreds and t is measured in months since the lake was stocked. After how many months will the fish population reach 500 fish? \n" ); document.write( "

Algebra.Com's Answer #776420 by ankor@dixie-net.com(22740) $\"\"$ $\"About$

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A small lake is stocked with a certain species of fish.
\n" ); document.write( "The fish population is modelled by the function $\"P%28t%29+=+10%2F%28%281%2B5e%5E%28-0.8t%29%29%29\"$ where
\n" ); document.write( " P is the number of fish in hundreds and
\n" ); document.write( " t is measured in months since the lake was stocked.
\n" ); document.write( "After how many months will the fish population reach 500 fish?
\n" ); document.write( ":
\n" ); document.write( "P is the number of fish in hundreds, therefore 500 fish is 5
\n" ); document.write( " $\"5+=+10%2F%28%281%2B5e%5E%28-0.8t%29%29%29\"$
\n" ); document.write( "multiply both sides by $\"%281%2B5e%5E%28-0.8t%29%29\"$
\n" ); document.write( " $\"5%281%2B5e%5E%28-0.8t%29%29+=+10\"$
\n" ); document.write( "divide both sides by 5
\n" ); document.write( " $\"1%2B5e%5E%28-0.8t%29+=+2\"$
\n" ); document.write( "subtract 1 from both sides
\n" ); document.write( " $\"5e%5E%28-0.8t%29+=+1\"$
\n" ); document.write( "divide both sides by 5
\n" ); document.write( " $\"e%5E%28-0.8t%29%29+=+.2\"$
\n" ); document.write( "using nat logs
\n" ); document.write( " $\"ln%28e%5E%28-.8t%29%29+=+ln%28.2%29\"$
\n" ); document.write( "-.8t*ln(e) = ln(.2)
\n" ); document.write( "ln of e is 1, find the ln of .2
\n" ); document.write( "-.8t = -1.6094
\n" ); document.write( "t = $\"%28-1.6094%29%2F-.8\"$
\n" ); document.write( "t ~ +2 months\r
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