document.write( "Question 837970: Steven absurd some red dye to turn himself red for Valentine's day. If the half-life of the dye is 2 hours, and there is only 20mg of dye left in him after 15 hours, what is the original amount of the dye he took? \n" ); document.write( "

Algebra.Com's Answer #504861 by KMST(5328) $\"\"$ $\"About$

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$\"y\"$ = amount of dye left in Steven, in mg, at time $\"t\"$ in hours.
\n" ); document.write( " $\"A\"$ = original amount of the dye, in mg, that he took, and had inside at $\"t=0\"$ .
\n" ); document.write( "
\n" ); document.write( "THE (precocious) FIFTH GRADER THOUGHTS:
\n" ); document.write( "Since the half-life of the dye is 2 hours,
\n" ); document.write( "after $\"red%281%29\"$ half-life, at $\"t%2F2=red%281%29\"$ , and
\n" ); document.write( " $\"y=A%2A%281%2F2%29\"$ is half of the original amount
\n" ); document.write( "After $\"red%282%29\"$ half-lives, $\"t%2F2=red%282%29\"$ another half-life has passed, and
\n" ); document.write( " $\"y=%28A%2A%281%2F2%29%29%2A%281%2F2%29=A%2A%281%2F2%29%5Ered%282%29\"$ .
\n" ); document.write( "Another half-life after that, after $\"red%283%29\"$ half-lives, $\"t%2F2=red%283%29\"$ , and
\n" ); document.write( " $\"y=%28A%2A%281%2F2%29%5E2%29%2A%281%2F2%29=A%2A%281%2F2%29%5Ered%283%29\"$ .
\n" ); document.write( "The pattern tells me that at $\"red%28t%2F2%29\"$ $\"half-lives\"$ the amount left is
\n" ); document.write( " $\"y=A%2A%281%2F2%29%5Ered%28t%2F2%29=A%2F2%5Ered%28%22t+%2F+2%22%29\"$ .
\n" ); document.write( "So if $\"20=A%2F2%5E%22t+%2F+2%22\"$ , $\"A=20%2A2%5E%22t+%2F+2%22\"$ .
\n" ); document.write( "Since $\"t=17\"$ , $\"t%2F2=17%2F2=8%261%2F2\"$ , and

$\"256%2A1.414\"$ .
\n" ); document.write( "A good approximate solution is
\n" ); document.write( " $\"A=20%2A256%2A1.414=7239.68=%22approx+.%22\"$ $\"highlight%287240%29\"$ .
\n" ); document.write( "Steven took in 7240 mg (7.24 grams) of red dye.
\n" ); document.write( "
\n" ); document.write( "THE PRE-CALCULUS STUDENT THOUGHTS:
\n" ); document.write( "I know the teacher said this type of problem is an example of exponential decay, and there is a formula for this.
\n" ); document.write( "It is an exponential function, with that irrational letter $\"e\"$ and a complicated exponent.
\n" ); document.write( "There was $\"t\"$ in that exponent, and something else, maybe the half-life, or some strange constant, or both.
\n" ); document.write( "
\n" ); document.write( "THE PRE-CALCULUS STUDENT WITH HELP FROM THE FIFTH GRADER:
\n" ); document.write( "I do mot enjoy memorizing and blindly applying formulas.
\n" ); document.write( "I keep forgetting those formulas, anyway.
\n" ); document.write( " $\"y=A%2F2%5Ered%28%22t+%2F+2%22%29\"$
\n" ); document.write( "That $\"red%28%22t+%2F+2%22%29\"$ is the number of half-lives,
\n" ); document.write( "because in this problem the half-life is $\"t%5B%221%2F2%22%5D=red%282%29\"$ ,
\n" ); document.write( "and so $\"t%2Ft%5B%221%2F2%22%5D=red%28t%2F2%29\"$ is the number of half-lives.
\n" ); document.write( " $\"ln%28y%29=ln%28A%2F2%5Ered%28%22t+%2F+2%22%29%29\"$
\n" ); document.write( " $\"ln%28y%29=ln%28A%29-ln%282%5Ered%28%22t+%2F+2%22%29%29\"$ --> $\"ln%28y%29%2Fln%28A%29=-red%28t%2F2%29%2Aln%282%29\"$ --> $\"ln%28y%29-ln%28A%29=-ln%282%29%2Ared%28t%2F2%29\"$ --> $\"ln%28y%2FA%29=-ln%282%29%2A%28t%2Ft%5B%221%2F2%22%5D%29\"$ --> $\"ln%28y%2FA%29=-ln%282%29%2At%2Ft%5B%221%2F2%22%5D\"$ --> $\"ln%28y%2FA%29=-%28ln%282%29%2Ft%5B%221%2F2%22%5D%29%2At\"$
\n" ); document.write( "I think that factor was called $\"lambda\"$ or $\"k\"$ ,
\n" ); document.write( " $\"lambda=ln%282%29%2Ft%5B%221%2F2%22%5D\"$ ,
\n" ); document.write( "and you can write
\n" ); document.write( " $\"ln%28y%2FA%29=-lambda%2At\"$ --> $\"y%2FA=e%5E%28-lambda%2At%29\"$ --> $\"y=A%2Ae%5E%28-lambda%2At%29\"$
\n" ); document.write( "Oh, and we were supposed to use the approximate value $\"ln%282%29=0.693\"$ ,
\n" ); document.write( "so $\"lambda=0.693%2F2-0.3465\"$ for this problem, so $\"y=A%2Ae%5E%28-0.3465%2At%29\"$ .
\n" ); document.write( "At $\"t=17\"$ , we have $\"20=A%2Ae%5E%28-0.3465%2A17%29\"$
\n" ); document.write( "Solving for $\"A\"$ ,
\n" ); document.write( " $\"A=20%2Fe%5E%28-0.3465%2A17%29=20%2Fe%5E%28-5.8905%29=20%2F0.00276559\"$ (rounding)
\n" ); document.write( "So $\"A=7232\"$ (rounded).
\n" ); document.write( "What about using a better approximation for $\"ln%282%29\"$ ?
\n" ); document.write( "With all the digits my calculator carries, $\"A=7241\"$ (rounded). \n" ); document.write( "

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