document.write( "Question 612930: find the interest rate needed for $6500 to grow to $10400 in 3 years if interest is continously compounded. \n" ); document.write( "
Algebra.Com's Answer #385800 by jim_thompson5910(35256)\"\" \"About 
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\"A=Pe%5E%28rt%29\" Start with the continuous compounding formula.\r
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\n" ); document.write( "\n" ); document.write( "\"10400=6500%2Ae%5E%28r%2A3%29\" Plug in \"A=10400\", \"P=6500\", and \"t=3\".\r
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\n" ); document.write( "\n" ); document.write( "\"10400%2F6500=e%5E%28r%2A3%29\" Divide both sides by \"6500\".\r
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\n" ); document.write( "\n" ); document.write( "\"1.6=e%5E%28r%2A3%29\" Evaluate \"10400%2F6500\" to get \"1.6\".\r
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\n" ); document.write( "\n" ); document.write( "\"ln%281.6%29=ln%28e%5E%28r%2A3%29%29\" Take the natural log of both sides.\r
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\n" ); document.write( "\n" ); document.write( "\"ln%281.6%29=r%2A3%2Aln%28e%29\" Pull down the exponent using the identity \"ln%28x%5Ey%29=y%2Aln%28x%29%29\".\r
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\n" ); document.write( "\n" ); document.write( "\"ln%281.6%29=r%2A3%2A1\" Evaluate the natural log of 'e' to get 1.\r
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\n" ); document.write( "\n" ); document.write( "\"ln%281.6%29=r%2A3\" Multiply and simplify.\r
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\n" ); document.write( "\n" ); document.write( "\"0.470003629245736=r%2A3\" Evaluate the natural log of \"1.6\" to get \"0.470003629245736\" (this value is approximate).\r
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\n" ); document.write( "\n" ); document.write( "\"0.470003629245736%2F3=r\" Divide both sides by \"3\" to isolate 'r'.\r
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\n" ); document.write( "\n" ); document.write( "\"0.156667876415245=r\" Evaluate \"0.470003629245736%2F3\" to get \"0.156667876415245\".\r
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\n" ); document.write( "\n" ); document.write( "\"r=0.156667876415245\" Flip the equation.\r
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\n" ); document.write( "\n" ); document.write( "\"r=0.1567\" Round to the nearest ten-thousandth.\r
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\n" ); document.write( "\n" ); document.write( "So the interest rate is roughly 15.67%
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