document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1174904>Question 1174904</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Find the sum of the ap:\r<BR>\n" );
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document.write( "logx(27/8), + logx(9/4), + logx(3/2)...(10 terms)\r<BR>\n" );
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document.write( " Answer is 15 (logx 2 − logx 3) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #800401 by </B><A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><B>math_tutor2020(3817)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=math_tutor2020><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=800401\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font color=black size=3><BR>\n" );
document.write( "Each log mentioned is base x\r<BR>\n" );
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document.write( "y = log(27/8) + log(9/4) + log(3/2) + ...  (summing 10 terms in this pattern)\r<BR>\n" );
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document.write( "y = log((3/2)^3) + log((3/2)^2) + log((3/2)^1) + ...\r<BR>\n" );
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document.write( "y = 3*log(3/2) + 2*log(3/2) + 1*log(3/2) + ...\r<BR>\n" );
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document.write( "y = (3+2+1+...)*log(3/2)\r<BR>\n" );
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document.write( "Now we must compute the arithmetic series 3+2+1+... all the way up to the tenth term of this.\r<BR>\n" );
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document.write( "We see the first term is a = 3 and the common difference is d = -1<BR>\n" );
document.write( "We want to sum n = 10 terms\r<BR>\n" );
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document.write( "S = (n/2)*(2*a + d(n-1))<BR>\n" );
document.write( "S = (10/2)*(2*3 - (10-1))<BR>\n" );
document.write( "S = -15<BR>\n" );
document.write( "The first ten terms of 3+2+1+... add up to -15\r<BR>\n" );
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document.write( "This means<BR>\n" );
document.write( "y = (3+2+1+...)*log(3/2)<BR>\n" );
document.write( "becomes<BR>\n" );
document.write( "y = -15*log(3/2)\r<BR>\n" );
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document.write( "Then we can use the log rule log(A/B) = log(A)-log(B) to get\r<BR>\n" );
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document.write( "y = -15*log(3/2)<BR>\n" );
document.write( "y = -15*(log(3) - log(2))<BR>\n" );
document.write( "y = 15*(-log(3) + log(2))<BR>\n" );
document.write( "y = 15*(log(2) - log(3))\r<BR>\n" );
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document.write( "Overall, this shows why <BR>\n" );
document.write( "log(27/8) + log(9/4) + log(3/2) + ...<BR>\n" );
document.write( "turns into<BR>\n" );
document.write( "15*(log(2) - log(3))\r<BR>\n" );
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document.write( "Each log mentioned in this solution is base x. To avoid dividing by zero, we cannot have x = 1. We also can't have <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x+%3C=+0 ALIGN=MIDDLE  ALT=\"x+%3C=+0\"   DISABLED_style= \"max-width:90%; height:auto;\" > if we want the output of the log to be a real number. <BR>\n" );
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