Question 80072
{{{A=A[0]e^(-0.058t)}}}
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Let's just assume that the original amount {{{A[0]}}} is 1.  Then to decay by half the amount
the resulting value of {{{A}}} would be {{{1/2}}}.  Substitute these two values and you
get:
.
{{{1/2=1*e^(-0.058t)}}}
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and after the multiplication by 1, the right side becomes just:
.
{{{1/2=e^(-0.058t)}}}
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Because an exponent contains the variable we are to solve for, that's a clue that we can use
logarithms and the rules of logarithms to solve.  And because e is also involved, let's
take the natural log of both sides to get:
.
{{{ln(1/2) = ln(e^(-0.058t))}}}
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A rule of logarithms that can be applied is that if the quantity that the logarithm
acts on has an exponent, you can make that exponent a multiplier of the logarithm.
Applying this rule to our problem results in:
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{{{ln(1/2) = -0.058t*ln(e)}}}
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But the natural logarithm of e is equal to 1.  So substitute 1 for {{{ln(e)}}} to get:
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{{{ln(1/2) = -0.058t * 1}}}
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Simplify the right side by doing the multiplication and get:
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{{{ln(1/2) = -0.058t}}}
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Now you can use a calculator to find the natural logarithm of {{{1/2}}} or its equivalent 0.5.
This logarithm is -0.69314718 and when this is substituted for {{{ln(1/2)}}} the equation
becomes:
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{{{ -0.69314718 = -0.058t}}}
.
Finally solve for t by dividing both sides by -0.058 to get:
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{{{t = (-0.69314718)/(-0.058) = 11.95081346}}}
.
Therefore, the closest answer in your list of possible answers is 12 days.  Hope this helps
you to see a way to solve problems such as these.