Question 72953
Given:
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{{{A = A[0]*e^(-0.058*t)}}}
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A is the amount of material at time t.  {{{A[0]}}} is the starting amount of material. And
t is the amount of elapsed time in days. You are asked to find the amount of time t that it
will take for the amount of material to equal half the amount you started with.
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Since you started with {{{A[0]}}} when it is half gone the A will equal {{{(1/2)* A[0]}}}
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Substituting this in for A results in the equation becoming:
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{{{(1/2)*A[0] = A[0]*e^(-0.058*t)}}}
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If you divide both sides of this equation by {{{A[0]}}} the {{{A[0]}}} term on both sides
drops out and you are left with:
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{{{(1/2) = e^(-0.058*t)}}}
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Take the logarithm of both sides.  Since the equation has an e in it, take the natural logarithm,
ln:
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ln(1/2) = ln(e^(-0.058*t))
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On a calculator you will find that ln(1/2) = ln(0.5) = -0.69314718.  Substitute this into
the equation for ln(1/2) to get:
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-0.69314718 = ln(e^(-0.058*t))
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On the right side, by the rules of logarithms in a logarithm of a term with an exponent
the exponent can be taken as the multiplier of the logarithm. As a result, the term:
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ln(e^(-0.058*t))
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is equal to the term:
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(-0.058*t)*ln(e)
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but ln(e) is equal to 1.  So this term further reduces to (-0.058*t).  Substitute this
into the equation and you now have:
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-0.69314718 = (-0.058*t)
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Multiply both sides by -1 to eliminate the negative signs and have:
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0.69314718 = 0.058*t
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Finally divide both sides by 0.058 to solve for t:
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0.69314718/0.058 = t
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And after the division is performed the equation becomes:
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11.9508 = t
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So after about 12 days one-half of the original material has decayed.
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Hope this helps you to decipher the problem.