document.write( "Question 557336: iF THE RATE OF DEPRECIATION IS APPROX. 30% how do I find the decay factor? \n" ); document.write( "
Algebra.Com's Answer #362588 by Theo(13342)\"\" \"About 
You can put this solution on YOUR website!
rate of depreciation per year is 30%.
\n" ); document.write( "that might also be equivalent to rate of decay.
\n" ); document.write( "b^x formula or you can use e^kx formula.
\n" ); document.write( "they will both provide you with the same answer.
\n" ); document.write( "b^x formula takes your value and raises it to an exponent for the number of time periods of your decay.
\n" ); document.write( "assume your original value is 100.
\n" ); document.write( "you have 30% depreciation / decay per year.
\n" ); document.write( "using the b^x factor, your formula would be:
\n" ); document.write( "f = p * (1 - decay factor) raised to the number of time periods.
\n" ); document.write( "with your numbers, that formula would be:
\n" ); document.write( "f = 100 * (.7)^n
\n" ); document.write( "if n = 0, then f = 100
\n" ); document.write( "if n = 1 then f = .70 * 100 = 70
\n" ); document.write( "if n = 2 then f = .70 * 70 = 49
\n" ); document.write( "etc.
\n" ); document.write( "using the e^kx factor, your formula would be:
\n" ); document.write( "first you would need to find the value of k.
\n" ); document.write( "you would do this by using a known decay and finding out what k is.
\n" ); document.write( "example:
\n" ); document.write( "you know that 30% decay after the first year results in 70
\n" ); document.write( "you would use this fact to find the value of k as follows:
\n" ); document.write( "70 = 100 * e^kx
\n" ); document.write( "e is the scientific constant of 2.718281828...
\n" ); document.write( "you would divide both sides of this equation by 100 to get:
\n" ); document.write( ".7 = e^(k*1)
\n" ); document.write( "you would then take the natural log of both sides of this equation to get:
\n" ); document.write( "ln(.7) = ln(e^k)
\n" ); document.write( "by the laws of logarithms, this becomes:
\n" ); document.write( "ln(.7) = k*ln(e)
\n" ); document.write( "since ln(e) = 1, this formula becomes:
\n" ); document.write( "ln(.7) = k
\n" ); document.write( "you would then find the natural log of .7 to get:
\n" ); document.write( "k = -.356674944
\n" ); document.write( "that's the value of k that would be used if you are using the e^kx formula.
\n" ); document.write( "let's see how both formulas work.
\n" ); document.write( "your starting value is 100
\n" ); document.write( "your decay factor is .3 per year.
\n" ); document.write( "you want to know the end value after 15 years.
\n" ); document.write( "using the b^x formula, you would do the following:
\n" ); document.write( "f = p * (.7)^15 which becomes:
\n" ); document.write( "f = 100 * (.7)^15 which becomes:
\n" ); document.write( "f = .474756151
\n" ); document.write( "using the e^kx formula, you would do the following:
\n" ); document.write( "f = p * e^kx
\n" ); document.write( "k = -.356674944
\n" ); document.write( "f = p * e^(-.356674944*15) which becomes:
\n" ); document.write( "f = 100 * e^(-.356674944*15) which becomes:
\n" ); document.write( "f = 100 * e^(-5.350124159) which becomes:
\n" ); document.write( "f = .474756151
\n" ); document.write( "you get the same answer either way.
\n" ); document.write( "the e^kx formula is used a lot in scientific studies.
\n" ); document.write( "the b^x formula, in this case, will provide the same answer as the e^kx formula.
\n" ); document.write( " \n" ); document.write( "
\n" );