document.write( "Question 1199046: A computer that was purchased for $6000 depreciates at a rate of 13.4% per year. An equation modeling this is, f(x)=6000(0.866)x, where x is the number of years.\r
\n" ); document.write( "\n" ); document.write( "What is the computer worth after 2 years? \r
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\n" ); document.write( "\n" ); document.write( "When will the value of the computer be $2530?
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Algebra.Com's Answer #832732 by Theo(13342)\"\" \"About 
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after two years, the computer will be worth 6000 * .866 * .866 = 6000 * .866^2 = 4499.736.
\n" ); document.write( "to find how long it takes for the computer to be equal o 2530, use the following formula.
\n" ); document.write( "2530 = 6000 * .866 ^ n
\n" ); document.write( "n is the number of years.
\n" ); document.write( "divide both sides of the equation by 6000 to get:
\n" ); document.write( "2530/6000 = .866 ^ n
\n" ); document.write( "take the log of both sides of the equation to get:
\n" ); document.write( "log(2530/6000) = log(.866 ^ n)
\n" ); document.write( "by log rules, this becomes:
\n" ); document.write( "log(2530/6000) = n * log(.866)
\n" ); document.write( "solve for n to get:
\n" ); document.write( "n = log(2530/6000) / log(.866) = 6.002209934.
\n" ); document.write( "confirm by replacing n in the original equation to get:
\n" ); document.write( "2530 = 6000 * .866 ^ 6.002209934.
\n" ); document.write( "this becomes 2530 = 2530, confirming that vallue of n is correct.
\n" ); document.write( "your solution is that the value of the computer will be 2530 in 6.002209934 years.
\n" ); document.write( "round as required.
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