document.write( "Question 745760: log(A)=log(A[0])+0.1tlog(.8) Solve for A as a function of t.\r
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Algebra.Com's Answer #453998 by KMST(5396)\"\" \"About 
You can put this solution on YOUR website!
I would start with properties of logarithms, like
\n" ); document.write( "\"b%2Alog%28%28a%29%29=log%28%28a%5Eb%29%29\" and \"log%28%28x%29%29%2Blog%28%28y%29%29=log%28%28xy%29%29\"
\n" ); document.write( "\"log%28%28A%29%29=log%28%28A%5B0%5D%29%29%2B0.1t%2Alog%28%280.8%29%29\"
\n" ); document.write( "\"log%28%28A%29%29=log%28%28A%5B0%5D%29%29%2Blog%28%280.8%5E%280.1t%29%29%29\"
\n" ); document.write( "\"log%28%28A%29%29=log%28%28A%5B0%5D%2A0.8%5E%280.1t%29%29%29\"
\n" ); document.write( "\"A=A%5B0%5D%2A0.8%5E%280.1t%29\"
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\n" ); document.write( "EXTRA:
\n" ); document.write( "\"A\" as a function of \"t\" is an exponential decay function.
\n" ); document.write( "The graph looks like this
\n" ); document.write( "\"graph%28300%2C300%2C-10%2C90%2C-15%2C135%2C100%2A0.8%5E%280.1x%29%29\"
\n" ); document.write( "If you start with a quantity \"A%5B0%5D\" and lose 20% of what you have every 10 hours, at time \"t=10\" hours you will have 80% of \"A%5B0%5D\" left
\n" ); document.write( "\"A=A%5B0%5D%2A0.8%5E%280.1%2A10%29=A%5B0%5D%2A0.8%5E1=A%5B0%5D%2A0.8\"
\n" ); document.write( "At time \"t=20\" hours you will have 80% of that 80% left
\n" ); document.write( "\"A=A%5B0%5D%2A0.8%5E%280.1%2A20%29=A%5B0%5D%2A0.8%5E2=A%5B0%5D%2A0.64\" , which is 64% of \"A%5B0%5D\"
\n" ); document.write( "At time \"t=30\" hours you will have 80% of 80% of 80% left
\n" ); document.write( "\"A=A%5B0%5D%2A0.8%5E%280.1%2A30%29=A%5B0%5D%2A0.8%5E3=A%5B0%5D%2A0.512\" , which is 51.2% of \"A%5B0%5D\"
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