Question 239904
If the equation is {{{A = A[0]e^(-0.0077x)}}}, then read on. If not then repost your question with a correct and clear equation, using parentheses generously.<br>
When a subscript of 0 is used, it is often used to indicate the initial or starting value for that value. In this case, {{{A[0]}}} represents the initial amount of the radioactive material. So we can rephrase the problem as: If {{{A[0] = 700}}}, after how many years will the amount, A, be 150? So we want to solve the equation:
{{{150 = 700e^(-0.0077x)}}}
To solve for variables in the exponent like this, we often use logarithms. But first we will isolate the e and its exponent by dividing both sides by 700:
{{{150/700 = e^(-0.0077x)}}}
The fraction on the left will reduce to:
{{{3/14 = e^(-0.0077x)}}}
Now we will use logarithms. We can use any logarithm which our calculator can handle. The "natural" choice (excuse the pun) is to use natural logarithms because of the e. But in case your calculator only does base 10 logarithms I will finish the solution using both:<ul><li>Natural logarithms.<ul><li>Find the natural logarithm of each side:
{{{ln(3/14) = ln(e^(-0.0077x))}}}</li><li>Use the property {{{log(a, (p^q)) = q*log(a, (p))}}} to move the exponent out in front:
{{{ln(3/14) = (-0.0077x)ln(e)}}}</li><li>Since ln(e) = 1:
{{{ln(3/14) = -0.0077x}}}</li><li>Divide both sides by -0.0077:
{{{ln(3/14)/(-0.0077) = x}}}</li><li>Use your calculator to simplify the left side
{{{(-1.5404450409471489)/(-0.0077) = x}}}
{{{200.0577975256037562 = x}}}</li></ul></li><li>Base 10 (aka Common) logarithms.<ul><li>Find the base 10 logarithm of each side:
{{{log((3/14)) = log((e^(-0.0077x)))}}}</li><li>Use the property {{{log(a, (p^q)) = q*log(a, (p))}}} to move the exponent out in front:
{{{log((3/14)) = (-0.0077x)log((e))}}}</li><li>Divide both sides by {{{(-0.0077*log((e)))}}}:
{{{log((3/14))/(-0.0077*log((e))) = x}}}</li><li>Substitute the approximation 2.7182818284590451 for e. (Feel free to round this off some.):
{{{log((3/14))/(-0.0077*log((2.7182818284590451))) = x}}}</li>
<li>Use your calculator to simplify the left side
{{{(-0.6690067809585756)/(-0.0077*0.4342944819032518) = x}}}
{{{(-0.6690067809585756)/(-0.0033440675106550) = x}}}
{{{200.0577975256037668 = x}}}</li></ul></li></ul>
So it will be about 200 years before the 700 pounds will be reduced to 150 pounds. (Note that we get the same answer using either natural or base 10 logarithms. (The very slight difference is due to using an approximation for e.))