document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=73568>Question 73568</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Could you please help me solve this interest word problem?  I'm clueless and don't know where to start! Here is the 2 part question.\r<BR>\n" );
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document.write( "The formula for calculating the amount of money returned for an initial deposit money into a bank account or CD is given by <BR>\n" );
document.write( "A=P( 1+r)^nt<BR>\n" );
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document.write( "       n\r<BR>\n" );
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document.write( "However, my question is now suppose, instead of knowing t, we know that the bank returned to us $15,000 with the bank compounding continuously. Using natural logarithms, find how long we have left the money in the bank (find t). Round your answer to the hundredth's place. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #52621 by </B><A HREF=/tutors/aboutme.mpl?userid=bucky><B>bucky(2189)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=bucky><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=52621\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Given:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+P%2A%281+%2B+%28r%2Fn%29%29%5E%28n%2At%29 ALIGN=MIDDLE  ALT=\"A+=+P%2A%281+%2B+%28r%2Fn%29%29%5E%28n%2At%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "where A is the amount returned, P is the initial deposit, n is the number of times the compounding<BR>\n" );
document.write( "takes place in a year, r is the annual interest rate expressed as a decimal, and t is the <BR>\n" );
document.write( "number of years the investment remains on deposit.<BR>\n" );
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document.write( "Actually, for continuous compounding, which is called for in the problem, the formula <BR>\n" );
document.write( "that should be used is:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+P%2Ae%5E%28r%2At%29 ALIGN=MIDDLE  ALT=\"A+=+P%2Ae%5E%28r%2At%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "where A, P, r, and t are as defined above and e is the base of the natural logarithms.<BR>\n" );
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document.write( "You picked the wrong equation for continuous compounding. The one you gave is more useful<BR>\n" );
document.write( "for a limited number of compoundings in a year ... twice a year, quarterly, monthly or<BR>\n" );
document.write( "weekly.  <BR>\n" );
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document.write( "So you need to proceed with the correct equation to use in solving for t subject to continuous<BR>\n" );
document.write( "compounding:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=A+=+P%2Ae%5E%28r%2At%29 ALIGN=MIDDLE  ALT=\"A+=+P%2Ae%5E%28r%2At%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "Now substitute $15,000 for A to get:<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+15000+=+P%2Ae%5E%28r%2At%29 ALIGN=MIDDLE  ALT=\"+15000+=+P%2Ae%5E%28r%2At%29\"   DISABLED_style= \"max-width:90%; height:auto;\" ><BR>\n" );
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document.write( "the fact that there is an exponent that contains the unknown suggests that we take the<BR>\n" );
document.write( "logarithm of both sides so we can use the logarithmic property that in a logarithm<BR>\n" );
document.write( "of a quantity having an exponent you can remove the exponent and use it as a multiplier<BR>\n" );
document.write( "of the logarithm. And since e (the base of natural logs) is involved in the equation,<BR>\n" );
document.write( "this suggests that we take the natural log of both sides of this equation (ln is the way<BR>\n" );
document.write( "we identify natural logarithms).<BR>\n" );
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document.write( "I haven't figured a way to make the formula translator at this site work on natural logs.  <BR>\n" );
document.write( "Therefore, I'm going back to the text way of writing equations.  Sorry ...\r<BR>\n" );
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document.write( "After taking the natural logarithm of both sides we have:<BR>\n" );
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document.write( "ln(15000) = ln(P*e^(rt))<BR>\n" );
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document.write( "A calculator tells you that ln(15000) = 9.61580548    Substitute this to give:<BR>\n" );
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document.write( "9.61580548 = ln(P*e^(rt))<BR>\n" );
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document.write( "By the rules of logarithms the log of a product is equal to the sums of the logs of the<BR>\n" );
document.write( "terms in the product.  Therefore we can write the right side of this equation as:<BR>\n" );
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document.write( "9.61580548 = ln(P) + ln(e^(r*t))<BR>\n" );
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document.write( "and as explained above, the exponent (rt) can be written as the multiplier of the log<BR>\n" );
document.write( "to give us:<BR>\n" );
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document.write( "9.61580548 = ln(P) + (r*t)*ln(e)<BR>\n" );
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document.write( "But ln(e) equals 1 and this reduces the problem to:<BR>\n" );
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document.write( "9.61580548 = ln(P) + r*t<BR>\n" );
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document.write( "ln(P) is just a number you can get from a calculator.  In solving for t you need it on<BR>\n" );
document.write( "the other side of the equation, so subtract ln(P) from both sides of the equation to<BR>\n" );
document.write( "give you:<BR>\n" );
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document.write( "9.61580548 - ln(P) = r*t<BR>\n" );
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document.write( "Now you can solve for t (in years) by dividing the entire left side of this equation by<BR>\n" );
document.write( "r.  I would help you with this, but you didn't provide P or r in your description of<BR>\n" );
document.write( "the problem.  (r should be something like .06 for a 6% annual interest rate.  Use the <BR>\n" );
document.write( "decimal form).  <BR>\n" );
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document.write( "Hope this helps you to get to a solution for this problem. </SPAN>\n" );
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