document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1198954>Question 1198954</A>: <I><SPAN STYLE=\"word-break: break-all;\"> b) The half-life of a radioactive substance is 300 years. If the initial amount is q milligrams, then the quantity q(t) remaining after t years is given by q(t) = q2^(kt). Find k </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #832644 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=832644\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "I'll use the variable 'a' as the initial amount, rather than q, since q is already taken in q(t).\r<BR>\n" );
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document.write( "Therefore, I'll rewrite this given equation<BR>\n" );
document.write( "q(t) = q*2^(kt)<BR>\n" );
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document.write( "q(t) = a*2^(kt)\r<BR>\n" );
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document.write( "initial amount = a<BR>\n" );
document.write( "half of that is q/2 = 0.5a<BR>\n" );
document.write( "which occurs t = 300 years later\r<BR>\n" );
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document.write( "This means q(t) = 0.5a when t = 300<BR>\n" );
document.write( "In other words, q(300) = 0.5a\r<BR>\n" );
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document.write( "q(t) = a*2^(kt)<BR>\n" );
document.write( "q(300) = a*2^(k*300)<BR>\n" );
document.write( "0.5a = a*2^(300k)<BR>\n" );
document.write( "0.5 = 2^(300k)\r<BR>\n" );
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document.write( "I divided both sides by 'a' in the jump from step 3 to step 4. <BR>\n" );
document.write( "Ultimately the starting amount does not matter.\r<BR>\n" );
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document.write( "From here we use logarithms to solve for k.<BR>\n" );
document.write( "If the variable is in the trees (aka exponent), then log it down.\r<BR>\n" );
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document.write( "The reason for using logs is because of this property<BR>\n" );
document.write( "log(A^B) = B*log(A)<BR>\n" );
document.write( "which helps us pull down the exponent to isolate the variable.\r<BR>\n" );
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document.write( "So,<BR>\n" );
document.write( "0.5 = 2^(300k)<BR>\n" );
document.write( "Log(0.5) = Log( 2^(300k) )<BR>\n" );
document.write( "Log(0.5) = 300k*Log( 2 )<BR>\n" );
document.write( "k = Log(0.5)/(300*Log(2))<BR>\n" );
document.write( "k = Log(2^(-1))/(300*Log(2))<BR>\n" );
document.write( "k = -1*Log(2)/(300*Log(2))<BR>\n" );
document.write( "<font color=red>k = -1/300</font>\r<BR>\n" );
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document.write( "Here's another way to solve for k without using logs.\r<BR>\n" );
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document.write( "0.5 = 2^(300k)<BR>\n" );
document.write( "1/2 = 2^(300k)<BR>\n" );
document.write( "2^(-1) = 2^(300k)\r<BR>\n" );
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document.write( "Both sides have the same base.<BR>\n" );
document.write( "In order for both sides to be equal, the exponents must be equal.\r<BR>\n" );
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document.write( "-1 = 300k<BR>\n" );
document.write( "300k = -1<BR>\n" );
document.write( "<font color=red>k = -1/300</font>\r<BR>\n" );
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document.write( "This type of trick is handy if you aren't too familiar with logs yet. <BR>\n" );
document.write( "However, this trick won't always work for any general exponential equation.\r<BR>\n" );
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document.write( "For example, the trick won't work if we had say<BR>\n" );
document.write( "0.7 = 2^(300k)<BR>\n" );
document.write( "which is why I prefer the logarithm approach.\r<BR>\n" );
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document.write( "Answer: <font color=red>k = -1/300</font>\r<BR>\n" );
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document.write( "-1/300 = -0.003333 where the '3's go on forever.<BR>\n" );
document.write( "I think it's best to stick to the fraction form since it's most exact. <BR>\n" );
document.write( "If your teacher says otherwise, then be sure to follow those instructions.<BR>\n" );
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