document.write( "Question 191254: My problem is with entering e raised to a negative power on my TI-84 Plus calculator. I know what answer I need to get and I can't figure out how to enter the problem correctly. I've looked on line and through my calculator manual to no avail!\r
\n" ); document.write( "\n" ); document.write( "The problem is: If 18 g of a radioactive substance are present initially and 3 yr later only 6 g remain, how much of the substance will be present after 11 yrs?\r
\n" ); document.write( "\n" ); document.write( "I get to the equation y=18e ^-.2310 x 7 and I know my answer is 5.02 grams, but how do I do this on my calc to get this answer? Thanks for any help! \n" ); document.write( "
Algebra.Com's Answer #143542 by Edwin McCravy(20056)\"\" \"About 
You can put this solution on YOUR website!
 \r\n" );
document.write( "\"A+=+Pe%5Ert\"\r\n" );
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document.write( "When \"t=0\", \"A=18\", plug those in:\r\n" );
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document.write( "\"A+=+Pe%5Ert\"\r\n" );
document.write( "\"18+=+Pe%5Er%280%29\"\r\n" );
document.write( "\"18+=+Pe%5E0\"\r\n" );
document.write( "\"18+=+P%281%29\"\r\n" );
document.write( "\"18+=+P\"\r\n" );
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document.write( "Substitute 18 for P in\r\n" );
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document.write( "\"A+=+18e%5Ert\"\r\n" );
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document.write( "When \"t=3\", \"A=6\", plug those in:\r\n" );
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document.write( "\"A+=+18e%5Ert\"\r\n" );
document.write( "\"6+=+18e%5Er%283%29\"\r\n" );
document.write( "\"6+=+18e%5E%283r%29\"\r\n" );
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document.write( "Divide both sides by 18\r\n" );
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document.write( "\"+6%2F18+=+%28+18%2Ae%5E%283r%29+%29%2F18+\"\r\n" );
document.write( "\"1%2F3+=+e%5E%283r%29\"\r\n" );
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document.write( "Use principle \"A=e%5EN\" is equivalent to \"N=ln%28A%29\"\r\n" );
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document.write( "\"3r=ln%281%2F3%29\"\r\n" );
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document.write( "Divide both sides by 3\r\n" );
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document.write( "\"r+=+ln%281%2F3%29%2F3\"\r\n" );
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document.write( "In your TI-84 type  ln(1/3)/3 ENTER and get -.3662040962\r\n" );
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document.write( "\"r+=+-.3662040962+\"\r\n" );
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document.write( "Substitute -.3662040962 for r in\r\n" );
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document.write( "\"A+=+18e%5E%28rt%29\"\r\n" );
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document.write( "\"A+=+18e%5E%28-.3662040962t%29\" \r\n" );
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document.write( "Now we substitute 11 for t\r\n" );
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document.write( "\"A+=+18e%5E%28-.3662040962t%29\" \r\n" );
document.write( "\"A+=+18e%5E%28-.3662040962%2811%29%29\"\r\n" );
document.write( "\"A+=+18e%5E%28-4.028245058%29\"\r\n" );
document.write( "\"A+=+18%28.0178055503%29\"\r\n" );
document.write( "\"A+=+.3204999045\"\r\n" );
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document.write( "So after 11 years, there are only .3205 grams.\r\n" );
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document.write( "Edwin
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