SOLUTION: I need help with this word problem: The radioactive bismuth isotope (210 Bi)disinigrates according to q=k(2)^(-t/5) where k is constant and t is the time in days. Express t in t

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Question 116789This question is from textbook Fund of Algo and trig
: I need help with this word problem:
The radioactive bismuth isotope (210 Bi)disinigrates according to q=k(2)^(-t/5) where k is constant and t is the time in days. Express t in terms of Q and k.
Can anyone solve or set up this problem?
thanks This question is from textbook Fund of Algo and trig

Answer by bucky(2189) About Me  (Show Source):
You can put this solution on YOUR website!
This is one way of working the problem.
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Given:
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q+=+k%2A%282%5E%28-t%2F5%29%29
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Begin by dividing both sides of this equation by k to get:
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q%2Fk+=+2%5E%28-t%2F5%29
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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%2810%2C%28q%2Fk%29%29+=+log%2810%2C2%5E%28-t%2F5%29%29
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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%2810%2C%28q%2Fk%29%29+=+%28-t%2F5%29%2Alog%2810%2C2%29
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Divide both sides by log%2810%2C2%29 and you get:
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%28log%2810%2C%28q%2Fk%29%29%29%2F%28log%2810%2C2%29%29+=+-t%2F5
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Multiply both sides of this equation by -5 to solve for t and you have:
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%28-5%2Alog%2810%2C%28q%2Fk%29%29%29%2F%28log%2810%2C2%29%29+=+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+=+%28-5%2Alog%2810%2C%28q%2Fk%29%29%29%2F%28log%2810%2C2%29%29
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You can simplify this a little further by recognizing that log%2810%2C2%29 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+=+%28-5%2Alog%2810%2C%28q%2Fk%29%29%29%2F%280.3010%29%29
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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%2Alog%2810%2C%28q%2Fk%29%29%29
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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%2Alog%2810%2C%28q%2Fk%29%29%29
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You could stop here or you could use the rule that log%2810%2C%28q%2Fk%29%29+=+log%2810%2Cq%29+-+log%2810%2Ck%29
and substitute into the equation for t to get:
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t+=+-16.6113%2A%28log%2810%2Cq%29-log%2810%2Ck%29%29
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Note that you can expect that log%2810%2Ck%29 must be bigger than log%2810%2Cq%29 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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