document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1078090>Question 1078090</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Suppose you deposit $1500 in a savings account that pays interest at an annual rate of 6%. No money is added or withdrawn from the account. How many years will it take for the account to contain $2500? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #692570 by </B><A HREF=/tutors/aboutme.mpl?userid=addingup><B>addingup(3677)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=addingup><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/cgi-bin/show-face.mpl?user=addingup><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=692570\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 2500 = 1500(1+0.06)^t<BR>\n" );
document.write( "So we have our unknown as a power function (^t). To solve, we need to use logs. <BR>\n" );
document.write( ":<BR>\n" );
document.write( "t = [log(A)-log(P)]/log(1+r) where A is the amount, P the principal, and r the rate.<BR>\n" );
document.write( "t = [log(2500)-log(1500)]/log(1.06) = 8.88, or 8.9 years.<BR>\n" );
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document.write( "Check:<BR>\n" );
document.write( "2500 = 1500(1+0.06)^8.8 Correct, check it out, comes out to 2504 due to rounding. </SPAN>\n" );
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