Question 116789
This is one way of working the problem.
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Given:
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{{{q = k*(2^(-t/5))}}}
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Begin by dividing both sides of this equation by k to get:
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{{{q/k = 2^(-t/5)}}}
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Take the logarithm of both sides. I chose to use base 10, but you could just as easily use
base e or any other base you prefer.
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{{{log(10,(q/k)) = log(10,2^(-t/5))}}}
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By the rules of logarithms, the exponent on the right side can be brought out as a multiplier
of the logarithm to give:
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{{{log(10,(q/k)) = (-t/5)*log(10,2)}}}
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Divide both sides by {{{log(10,2)}}} and you get:
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{{{(log(10,(q/k)))/(log(10,2)) = -t/5}}}
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Multiply both sides of this equation by -5 to solve for t and you have:
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{{{(-5*log(10,(q/k)))/(log(10,2)) = t}}}
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Just to get it into a little more standard form, transpose the equation (switch sides) and
you have:
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{{{t = (-5*log(10,(q/k)))/(log(10,2))}}}
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You can simplify this a little further by recognizing that {{{log(10,2)}}} is just a 
number. A calculator will tell you that it is 0.301028885 and if you like, you can round this
to 0.3010 (Choose the amount of rounding you want.) This makes the equation:
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{{{t = (-5*log(10,(q/k)))/(0.3010))}}}
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and if you divide the denominator into the -5 in the numerator, the equation reduces to:
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{{{t = -16.61129568*log(10,(q/k)))}}}
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Might as well round off -16.61129568 to -16.6113 or whatever you want ... to get:
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{{{t = -16.6113*log(10,(q/k)))}}}
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You could stop here or you could use the rule that {{{log(10,(q/k)) = log(10,q) - log(10,k)}}}
and substitute into the equation for t to get:
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{{{t = -16.6113*(log(10,q)-log(10,k))}}}
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Note that you can expect that {{{log(10,k)}}} must be bigger than {{{log(10,q)}}} to prevent 
you from getting a negative answer for t.
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Check the above for errors. The basic process of using the logarithm is one that can be used
to solve this problem, though.
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Hope this helps you to get on with the problem.
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