document.write( "Question 1129782: I've been stuck trying to solve this. Could someone help explain how to correctly set up the equation. Thanks
\n" ); document.write( " A scientist begins with 250 grams of a radioactive substance. After 250 minutes, the sample has decayed to 32 grams. Write an exponential equation
\n" ); document.write( "f(t) representing this situation. (Let f be the amount of radioactive substance in grams and t be the time in minutes.) \r
\n" ); document.write( "\n" ); document.write( "The exponential equation that I thought was correct, but isn't is 250* 2( -t/h).
\n" ); document.write( " \n" ); document.write( "
Algebra.Com's Answer #746409 by MathTherapy(10552)\"\" \"About 
You can put this solution on YOUR website!
I've been stuck trying to solve this. Could someone help explain how to correctly set up the equation. Thanks
\n" ); document.write( " A scientist begins with 250 grams of a radioactive substance. After 250 minutes, the sample has decayed to 32 grams. Write an exponential equation
\n" ); document.write( "f(t) representing this situation. (Let f be the amount of radioactive substance in grams and t be the time in minutes.) \r
\n" ); document.write( "\n" ); document.write( "The exponential equation that I thought was correct, but isn't is 250* 2( -t/h).
\n" ); document.write( "
Exponential GROWTH rate formula: \"matrix%281%2C3%2C+y%2C+%22=%22%2C+A%5Bo%5Db%5Et%29\", where:
\n" );
document.write( "y   = REMAINING amount of substance (32, in this case)
\n" );
document.write( "\"A%5Bo%5D\" = INITIAL amount of substance (250 g, in this case)
\n" );
document.write( "b   = GROWTH rate of substance (Unknown, in this case)
\n" );
document.write( "t   = time substance takes to grow (250 minutes, in this case)
\n" );
document.write( "\"matrix%281%2C3%2C+f%28x%29%2C+%22=%22%2C+A%5Bo%5Db%5Et%29\"
\n" );
document.write( "\"matrix%281%2C3%2C+32%2C+%22=%22%2C+250b%5E250%29\" ------- Substituting 32 for f(x), or y, 250 for \"A%5Bo%5D\", and 250 for t
\n" );
document.write( "\"matrix%281%2C3%2C+32%2F250%2C+%22=%22%2C+b%5E250%29\"
\n" );
document.write( "\"matrix%281%2C3%2C+16%2F125%2C+%22=%22%2C+b%5E250%29\" ------- Reducing fraction on left
\n" );
document.write( " ------- Multiplying both sides by the \"matrix%281%2C2%2C+%281%2F250%29%5Eth%2C+root%29\"
\n" );
document.write( "
\n" );
document.write( "0.991810816, or .9918 = b
\n" );
document.write( "Exponential equation: \"highlight_green%28matrix%281%2C3%2C+f%28t%29%2C+%22=%22%2C+250%28.9918%29%5Et%29%29\"
\n" );
document.write( "Note: \"b\", being less than 1 (.9918) indicates a DECAY. If \"b\" were greater than 1, then that'd signify a GROWTH, as opposed to DECAY.\r
\n" );
document.write( "\n" );
document.write( "OR\r
\n" );
document.write( "\n" );
document.write( "As this is substance, which DECAYS/GROWS exponentially, the CONTINUOUS GROWTH/DECAY formula, or \"matrix%281%2C3%2C+A%2C+%22=%22%2C+A%5Bo%5De%5E%28kt%29%29\" is used.
\n" );
document.write( "\"matrix%281%2C3%2C+32%2C+%22=%22%2C+250e%5E%28250k%29%29\" -------- Substituting 32 for A, 250 for \"A%5Bo%5D\", and 250 for t
\n" );
document.write( "\"matrix%281%2C3%2C+32%2F250%2C+%22=%22%2C+e%5E%28250k%29%29\"
\n" );
document.write( "\"matrix%281%2C3%2C+16%2F125%2C+%22=%22%2C+e%5E%28250k%29%29\" ---------- Reducing fraction on left side
\n" );
document.write( "\"matrix%281%2C3%2C+250k%2C+%22=%22%2C+ln+%2816%2F125%29%29\" -------- Converting to NATURAL LOGARITHMIC (ln) form
\n" );
document.write( "k, or rate of growth/decay = \"matrix%281%2C3%2C+ln+%2816%2F125%29%2F250%2C+%22=%22%2C+-+0.0082229%29\"
\n" );
document.write( "As k is negative (< 0), then the substance DECAYED as opposed to it GROWING. This we already know based on the fact that initially the substance was 250 g, and reduced to 32 g in  250 hours. 
\n" );
document.write( "However, it's good to know that when the CONTINUOUS GROWTH/DECAY formula is used, GROWTH is indicated by a POSITIVE (> 0) value for k, and DECAY, by a NEGATIVE (< 0) value for k.\r
\n" );
document.write( "\n" );
document.write( "The equation for CONTINUOUS GROWTH/DECAY of a substance, or:  \n" );
document.write( "
\n" );