Question 70663
{{{p = 80e^(-q/2)}}}  Solve for q.
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This is what I understand your problem to be. Assuming this is correct, you can take the ln
(natural logarithm which has the base e) of both sides and the problem becomes:
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{{{ ln(p)= ln(80*e^(-q/2))}}}
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But the logarithm of a product is equals the sum of the logarithms of the two terms being
multiplied.  Therefore, we can split the right side into the sum of two logarithms as follows:
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{{{ ln(p)= ln(80)+ ln(e^(-q/2))}}}
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Subtract ln(80) from both sides to get:
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{{{ ln(p) - ln(80) =  + ln(e^(-q/2))}}}
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By the rules of logarithms, the difference of the logarithms of two quantities can be 
re-written as the logarithm of the quotients of the quantities.  This translates to:
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{{{ ln(p) - ln(80) = ln(p/80)}}}
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Substituting this as a replacement for the left side results in:
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{{{ ln(p/80) = ln(e^(-q/2))}}}
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Then by a rule of exponents in logarithms, the exponent of a term in a logarithm becomes the
multiplier of the logarithm of the term on the right side.  In this case {{{(-q/2)}}}
becomes the multiplier of ln(e) and the right side of the equation is changed as shown below:
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{{{ ln(p/80) = (-q/2)*ln(e)}}}
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But ln(e) = 1, and when this substitution is made the equation becomes:
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{{{ ln(p/80) = (-q/2)}}}
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Multiply both sides of the equation by -2 and the equation becomes:
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{{{ -2*ln(p/80) = q}}}
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This is the answer, but there is one additional constraint. The value of p must be greater 
than zero or else you would be taking the ln of a negative number or zero and those are 
outside of the allowed values of numbers that the ln function can operate on.
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Hopes this gives you some additional insight about the subject of logarithms and natural
logarithms in particular.