document.write( "Question 173403This question is from textbook introduction to college mathematics
\n" ); document.write( ": Use the exponential decay function P(t) = Poe^kt
\n" ); document.write( "Chemistry. The decay rate of krypton-85 is 6.3% per day. What is the half-life? \n" ); document.write( "
Algebra.Com's Answer #128281 by gonzo(654)\"\" \"About 
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the formula is:
\n" ); document.write( "\"p%5Bt%5D+=+p%5B0%5D%2Ae%5E%28-k%2At%29\"
\n" ); document.write( "that minus k is important since the exponent has to be negative in a decay problem.
\n" ); document.write( "-----
\n" ); document.write( "you are given:
\n" ); document.write( "decay rate is 6.3% per day.
\n" ); document.write( "this translates to .063 per day which is equal to k since k, in this case, is the decay rate per day.
\n" ); document.write( "-----
\n" ); document.write( "you are also asked for the half life.
\n" ); document.write( "this means at what point in time will the amount of krypton be half of what it is today.
\n" ); document.write( "if \"p%5B0%5D\" = 1 = the amount of krypton today, then \"p%5Bt%5D\" = .5 is the amount of krypton when one half life has occurred.
\n" ); document.write( "-----
\n" ); document.write( "your equation:
\n" ); document.write( "\"p%5Bt%5D+=+p%5B0%5D%2Ae%5E%28-k%2At%29\"
\n" ); document.write( "becomes:
\n" ); document.write( "\".5+=+1+%2A+e%5E%28-.063%2At%29\"
\n" ); document.write( "which becomes:
\n" ); document.write( "\".5+=+e%5E%28-.063%2At%29\"
\n" ); document.write( "which you can solve using the log or ln function of your calculator.
\n" ); document.write( "we'll use the log function.
\n" ); document.write( "either one will get the same answer.
\n" ); document.write( "-----
\n" ); document.write( "your equation becomes:
\n" ); document.write( "\"log%2810%2C.5%29+=+log%2810%2Ce%5E%28-.063%2At%29%29\"
\n" ); document.write( "since \"log%2810%2Cb%5Ea%29+=+a%2Alog%2810%2Cb%29\" your equation becomes:
\n" ); document.write( "\"log%2810%2C.5%29+=+-.063%2At%2Alog%2810%2Ce%29\"
\n" ); document.write( "if you divide both sides of this equation by \"-.063%2Alog%2810%2Ce%29\", you get:
\n" ); document.write( "\"log%2810%2C.5%29%2F%28-.063%2Alog%2810%2Ce%29%29+=+t\"
\n" ); document.write( "which you can solve using your calculator to get:
\n" ); document.write( "t = 11.0023362 days
\n" ); document.write( "-----
\n" ); document.write( "to prove this is true, plug that value in your original equation for t.
\n" ); document.write( "that equation:
\n" ); document.write( "\"p%5Bt%5D+=+p%5B0%5D%2Ae%5E%28-k%2At%29\"
\n" ); document.write( "becomes:
\n" ); document.write( "\".5+=+1%2Ae%5E%28-.063%2A11.0023362%29\"
\n" ); document.write( "which becomes:
\n" ); document.write( "\".5+=+e%5E%28-.693145181%29\"
\n" ); document.write( "which becomes:
\n" ); document.write( ".5 = .5
\n" ); document.write( "-----
\n" ); document.write( "value is good.
\n" ); document.write( "answer is:
\n" ); document.write( "t = 11.0023362 \n" ); document.write( "
\n" );