document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=31311>Question 31311</A>: <I><SPAN STYLE=\"word-break: break-all;\"> For questions 3-4, write the logarithmic equation in exponential form. For example, the exponential form of \"log(base 5)25 = 2\" is \"5^2 = 25\".\r<BR>\n" );
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document.write( "3.	(2 points)   log (base 3) 27 = 3 <BR>\n" );
document.write( "4.	(2 points)   log {base 125) 25 = 2/3<BR>\n" );
document.write( "For questions 5 and 6, recall that, when interest is compounded continuously, the balance in an account after t years is given by <BR>\n" );
document.write( "A = Pe^(rt)<BR>\n" );
document.write( "where P is the initial investment and r is the interest rate. \r<BR>\n" );
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document.write( "5.	(5 points)   Maya has deposited $600 in an account that pays 5.64% interest, compounded continuously. How long will it take for her money to double? \r<BR>\n" );
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document.write( "6.	(5 points)   Suppose that $2000 is invested at a rate of 6% per year compounded continuously. What is the balance after 1 yr? After 2 yrs?<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #18066 by </B><A HREF=/tutors/aboutme.mpl?userid=stanbon><B>stanbon(75887)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=stanbon><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=18066\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 3) 3^3=27<BR>\n" );
document.write( "4) 125^(2/3)=25\r<BR>\n" );
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document.write( "5)2(600)=600e^(0.0564t)<BR>\n" );
document.write( "  2= e^(0.0564t)<BR>\n" );
document.write( " Take the natural log of both sides to get:<BR>\n" );
document.write( " ln 2 = 0.0564t<BR>\n" );
document.write( "0.693147...=0.0564t<BR>\n" );
document.write( "t=12.2898.. years\r<BR>\n" );
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document.write( "6) A=2000e^(0.06) = 2000(1.062)=$2123.67 after one year\r<BR>\n" );
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document.write( "   A=2000e^(0.06(2))=2000e^(0.12)=$2254.99 after two years\r<BR>\n" );
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document.write( "Cheers,<BR>\n" );
document.write( "Stan H.\r<BR>\n" );
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