document.write( "Question 213724: Recall that the formula for compounding interest is A = P(1+ r)t where A is the total accumulated amount in an account with principal P compounded at a annual interest rate r once per year for t years.\r
\n" ); document.write( "\n" ); document.write( "a. How many years will it take for a principal of
\n" ); document.write( "P =1000 to reach an amount of A = 5000 when compounded once a year at an annual interest rate of 6%?\r
\n" ); document.write( "\n" ); document.write( "b. If the interest rate is doubled to 12%, what will the number of years be for the
\n" ); document.write( "principal of P =1000 to reach an amount of A = 5000?
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Algebra.Com's Answer #161453 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
a)\r
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\n" ); document.write( "\n" ); document.write( "\"A=P%281%2Br%29%5Et\" Start with the given equation.\r
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\n" ); document.write( "\n" ); document.write( "\"5000=1000%281%2B0.06%29%5Et\" Plug in \"P=1000\", \"A=5000\", and \"r=0.06\"\r
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\n" ); document.write( "\n" ); document.write( "\"5000=1000%281.06%29%5Et\" Add\r
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\n" ); document.write( "\n" ); document.write( "\"5000%2F1000=%281.06%29%5Et\" Divide both sides by 1000.
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\n" ); document.write( "\n" ); document.write( "\"5=%281.06%29%5Et\" Divide\r
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\n" ); document.write( "\n" ); document.write( "\"log%2810%2C%285%29%29=log%2810%2C%281.06%5Et%29%29\" Take the log of both sides.\r
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\n" ); document.write( "\n" ); document.write( "\"log%2810%2C%285%29%29=t%2Alog%2810%2C%281.06%29%29\" Pull down the exponent.\r
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\n" ); document.write( "\n" ); document.write( "\"0.69897=t%2Alog%2810%2C%281.06%29%29\" Evaluate the left side\r
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\n" ); document.write( "\n" ); document.write( "\"0.69897=t%2A0.02531\" Evaluate the log on the right side\r
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\n" ); document.write( "\n" ); document.write( "\"0.69897%2F0.02531=t\" Divide both sides by 0.02531.\r
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\n" ); document.write( "\n" ); document.write( "\"t=27.61636\" Divide and rearrange the equation.\r
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\n" ); document.write( "\n" ); document.write( "So it will take about 27.61 years (a little over 27 and a half years) for $1,000 to become $5,000 at an interest rate of 6%\r
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\n" ); document.write( "\n" ); document.write( "b)\r
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\n" ); document.write( "\n" ); document.write( "\"A=P%281%2Br%29%5Et\" Start with the given equation.\r
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\n" ); document.write( "\n" ); document.write( "\"5000=1000%281%2B0.12%29%5Et\" Plug in \"P=1000\", \"A=5000\", and \"r=0.12\"\r
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\n" ); document.write( "\n" ); document.write( "\"5000=1000%281.12%29%5Et\" Add\r
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\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"5000%2F1000=%281.12%29%5Et\" Divide both sides by 1000.
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\n" ); document.write( "\n" ); document.write( "\"5=%281.12%29%5Et\" Divide\r
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\n" ); document.write( "\n" ); document.write( "\"log%2810%2C%285%29%29=log%2810%2C%281.12%5Et%29%29\" Take the log of both sides.\r
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\n" ); document.write( "
\n" ); document.write( "\n" ); document.write( "\"log%2810%2C%285%29%29=t%2Alog%2810%2C%281.12%29%29\" Pull down the exponent.\r
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\n" ); document.write( "\n" ); document.write( "\"0.69897=t%2Alog%2810%2C%281.12%29%29\" Evaluate the left side\r
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\n" ); document.write( "\n" ); document.write( "\"0.69897=t%2A0.04922\" Evaluate the log on the right side\r
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\n" ); document.write( "\n" ); document.write( "\"0.69897%2F0.04922=t\" Divide both sides by 0.04922.\r
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\n" ); document.write( "\n" ); document.write( "\"t=14.20093\" Divide and rearrange the equation.\r
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\n" ); document.write( "\n" ); document.write( "So it will take about 14.2 years (a little over 14 years) for $1,000 to become $5,000 at an interest rate of 12%
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