document.write( "Question 718152: Please help me solve for A and B in this exponential decay model y= ae^-bx. I have the following points: when x=0.6, y=1000 and when x=1.8, y=1.\r
\n" ); document.write( "\n" ); document.write( "I have done the following but I am not convienced it is the right path...\r
\n" ); document.write( "\n" ); document.write( "y=ae-bx therefor e-bx=y/a ... take log e of both sides ... loge e-bx = loge y/a\r
\n" ); document.write( "\n" ); document.write( "but logee=1 therefor -bx=loge y/a ... move the x over to get -b=x loge y/a\r
\n" ); document.write( "\n" ); document.write( "then enter values for x and y from example 1... -b= (0.6) loge(1000/a)\r
\n" ); document.write( "\n" ); document.write( "Is this the correct path to take for solving this?? Any advice would be greafull appreciated. Thank you. \n" ); document.write( "
Algebra.Com's Answer #853669 by MathTherapy(10704)\"\" \"About 
You can put this solution on YOUR website!
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document.write( "Please help me solve for A and B in this exponential decay model y= ae^-bx. I have the following points: when x=0.6, y=1000 and when x=1.8, y=1.\r\n" );
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document.write( "I have done the following but I am not convienced it is the right path...\r\n" );
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document.write( "y=ae-bx therefor e-bx=y/a ... take log e of both sides ... loge e-bx = loge y/a\r\n" );
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document.write( "but logee=1 therefor -bx=loge y/a ... move the x over to get -b=x loge y/a\r\n" );
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document.write( "then enter values for x and y from example 1... -b= (0.6) loge(1000/a)\r\n" );
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document.write( "Is this the correct path to take for solving this?? Any advice would be greafull appreciated. Thank you.\r\n" );
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document.write( "I've chosen to show you the CORRECT path, instead of looking through your work to identify your error(s), if any!\r\n" );
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document.write( "       \"y+=+ae%5E%28-+bx%29\", if: \"system%28matrix%282%2C1%2C+x+=+0.6%2C+y+=+%221%2C000%22%29%29\"           \"y+=+ae%5E%28-bx%29\", if: \"system%28matrix%282%2C1%2C+x+=+1.8%2C++y+=+1%29%29\"\r\n" );
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document.write( "Equating the 2 \"a\" values, we get: \"%221%2C000%22%2Fe%5E%28-+.6b%29+=+1%2Fe%5E%28-+1.8b%29\"\r\n" );
document.write( "                         \"%221%2C000%22%28e%5E%28-+1.8b%29%29+=+e%5E%28-+.6b%29\" ---- Cross-multiplying\r\n" );
document.write( "                         \"%221%2C000%22%28e%5E%28-+1.8b%29%29%2Fe%5E%28-+1.8b%29+=+e%5E%28-+.6b%29%2Fe%5E%28-+1.8b%29\" --- Dividing both sides by \"e%5E%28-+1.8b%29\"\r\n" );
document.write( "                         \r\n" );
document.write( "                                    \"%221%2C000%22+=+e%5E%28-+.6b+-+%28-+1.8b%29%29\"\r\n" );
document.write( "                                    \"%221%2C000%22+=+e%5E%28-+.6b+%2B+1.8b%29\"\r\n" );
document.write( "                                    \"%221%2C000%22+=+e%5E%281.2b%29\"\r\n" );
document.write( "                                      1.2b = ln (1,000) --- Converting to LOGARITHMIC (natural) form \r\n" );
document.write( "                                        \"highlight%28b%29+=+%28ln+%28%221%2C000%22%29%29%2F1.2\" = 5.76, approximately.\r\n" );
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document.write( "You can now substitute this value for b in either \"a+=+%221%2C000%22%2Fe%5E%28-+.6b%29\", or \"a+=+1%2Fe%5E%28-+1.8b%29\" to determine the value of \"a.\"\r\n" );
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document.write( "If/When you do, you should get the value of a as approximately 31,622.8, to the nearest tenth.\r\n" );
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document.write( "@GREENESTAMPS is CORRECT! An INCORRECT value of 51,622.8 for \"a\" was entered previously, although the \r\n" );
document.write( "correct value, 31,622.8 was calculated. This has now been corrected. Thx for pointing out the error.
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