Question 214806
{{{B(t)=2000*e^(-.03*t)}}}
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Substitute 50 for B(t) to get:
{{{50 = 2000*e^(-.03*t)}}}
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Take the natural log of both sides to get:
{{{ln(50) = ln(2000*e^(-.03*t))}}}
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rules of logarithms apply here.
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first rule:
{{{ln(a*b) = ln(a) + ln(b)}}}
second rule:
{{{ln(a^b) = b*ln(a)}}}
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applying the first rule,
{{{ln(2000*e^(-.03*t))}}} becomes {{{ln(2000) + ln(e^(-.03t))}}}
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applying the second rule,
{{{ln(e^(-.03t))}}} becomes {{{-.03t*ln(e)}}}
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putting this all together, your equation becomes:
{{{ln(50) = ln(2000) - .03t*ln(e)}}}
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{{{ln(e)}}} equals 1, so your equation becomes:
{{{ln(50) = ln(2000) - .03t}}}
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subtract ln(2000) from both sides of this equation and then divide both sides of this equation by (-.03) to get:
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{{{(ln(50) - ln(2000))/(-.03) = t}}}
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this becomes:
{{{(3.912023005 - 7.60090246)/(-.03) = t}}}
this becomes:
{{{122.9626485 = t}}}
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The population will decay to 50 in 122.9626485 hours.
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substitute this value for t in the original equation to get:
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{{{B(t)=2000*e^(-.03*(122.9626485))}}}
this becomes:
{{{B(t) = 50}}} confirming that {{{t = 122.9626485}}} is a good value for t.
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