document.write( "Question 1201501: At time=0, there are 6,000 grams of a radioactive material present. The half-life of the element is 18 years. In how many years will there be 115 grams remaining? Round your answer to the nearest 0.01 years.\r
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Algebra.Com's Answer #836079 by MathTherapy(10552)\"\" \"About 
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document.write( "At time=0, there are 6,000 grams of a radioactive material present. The half-life of the element is 18 years. In how many years will there be 115 grams remaining? Round your answer to the nearest 0.01 years.\r\n" );
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document.write( "Any help would be appreciated, I have tried many problems like this one without success. \r\n" );
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document.write( "If ½-life is “a” time-periods, then k, or DECAY CONSTANT = \r\n" );
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document.write( "CONTINUOUS GROWTH/DECAY formula: \"matrix%281%2C3%2C+A%2C+%22=%22%2C+A%5Bo%5De%5E%28kt%29%29\", with:\r\n" );
document.write( "\"A\"  being remaining amount after time t (115, in this case)\r\n" );
document.write( "\"A%5Bo%5D\" being Original/Initial amount (6,000, in this case)\r\n" );
document.write( "\"k\"  being the constant (k > 0 signifies RATE OF GROWTH ;  k < 0 signifies RATE OF DECAY ; k = - .0385, in this case) \r\n" );
document.write( "\"t\"  being time, in stated periods (Unknown, in this case)\r\n" );
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document.write( "                                          \"matrix%281%2C3%2C+A%2C+%22=%22%2C+A%5Bo%5De%5E%28kt%29%29\"\r\n" );
document.write( "                                        \"matrix%281%2C3%2C+115%2C+%22=%22%2C+%226%2C000%22e%5E%28-+.0385t%29%29\" ----- Substituting 115 for A, 6,000 for \"A%5Bo%5D\", and - .0385 for k\r\n" );
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document.write( "                                   \"matrix%281%2C3%2C+-+.0385t%2C+%22=%22%2C+ln+%2823%2F%221%2C200%22%29%29\" ------ Converting to LOGARITHMIC (Natural) form\r\n" );
document.write( "Time it takes for 115 grams to remain, or \r\n" );
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document.write( "The correct answer should actually be in WHOLE-NUMBER years (103 to be specific)!
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