document.write( "Question 1087136: In what time will rupees 1000 becomes Rs. 1331 at 10% rate p.a compound interest. \n" ); document.write( "
Algebra.Com's Answer #701415 by Theo(13342)\"\" \"About 
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f = p * (1 + r) ^ n\r
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\n" ); document.write( "\n" ); document.write( "f is the future value
\n" ); document.write( "p is the present value
\n" ); document.write( "r is the interest rate per time period
\n" ); document.write( "n is the number of time periods.\r
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\n" ); document.write( "\n" ); document.write( "in your problem:\r
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\n" ); document.write( "\n" ); document.write( "f = 1331
\n" ); document.write( "p = 1000
\n" ); document.write( "r = .1 per year
\n" ); document.write( "n = number of years\r
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\n" ); document.write( "\n" ); document.write( "the formula becomes:\r
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\n" ); document.write( "\n" ); document.write( "1331 = 1000 * (1 + .1) ^ n\r
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\n" ); document.write( "\n" ); document.write( "divide both sides of the equation by 1000 and simplify to get:\r
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\n" ); document.write( "\n" ); document.write( "1.331 = 1.1 ^ n\r
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\n" ); document.write( "\n" ); document.write( "take the log of both sides of the equation to get:\r
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\n" ); document.write( "\n" ); document.write( "log(1.331) = log(1.1 ^ n)\r
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\n" ); document.write( "\n" ); document.write( "simplify to get log(1.331) = n * log(1.1)\r
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\n" ); document.write( "\n" ); document.write( "solve for n to get n = log(1.331) / log(1.1) = 3\r
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\n" ); document.write( "\n" ); document.write( "1000 rupees will become 1331 rupees in 3 years at an interest rate of 10% compounded annually.\r
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\n" ); document.write( "\n" ); document.write( "in general, log(b^n) = n*log(b)\r
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\n" ); document.write( "\n" ); document.write( "that's one of the basic properties of logarithms.\r
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