document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=289496>Question 289496</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Hi there. I need to obtain y in terms of x for the following logarithmic equation:\r<BR>\n" );
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document.write( "ln(y-1) = 2 ln x + ln y-x\r<BR>\n" );
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document.write( "The book answer (given without a working) is y = 1 / (1 - x^2 * E^-x),<BR>\n" );
document.write( "but using the logarithmic identities I can only seem to get as far as<BR>\n" );
document.write( "y = 1 + x^2 + yE^-x. I'm not sure if my error is in the log identities or the basic algebra. Thanks! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #209708 by </B><A HREF=/tutors/aboutme.mpl?userid=CharlesG2><B>CharlesG2(834)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CharlesG2><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=209708\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Hi there. I need to obtain y in terms of x for the following logarithmic equation: <BR>\n" );
document.write( "ln(y-1) = 2 ln x + ln y-x <BR>\n" );
document.write( "The book answer (given without a working) is y = 1 / (1 - x^2 * E^-x),<BR>\n" );
document.write( "but using the logarithmic identities I can only seem to get as far as<BR>\n" );
document.write( "y = 1 + x^2 + yE^-x. I'm not sure if my error is in the log identities or the basic algebra. Thanks!\r<BR>\n" );
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document.write( "ok, let me try\r<BR>\n" );
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document.write( "ln(y - 1) = 2ln(x) + ln(y-x)\r<BR>\n" );
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document.write( "ln is the natural log, it is base e, where e is Euler's Number approx.= 2.7183<BR>\n" );
document.write( "base b = e (by definition)<BR>\n" );
document.write( "ln(mn) = ln(m) + ln(n) (logarithmic rule)<BR>\n" );
document.write( "ln(m/n) = ln(m) - ln(n) (logarithmic rule)<BR>\n" );
document.write( "ln(n) = m --> e^m = n (logarithmic rule)<BR>\n" );
document.write( "ln(m^n) = n * ln(m) (logarithmic rule)<BR>\n" );
document.write( "e^(ln(m)) = m (logarithmic rule)<BR>\n" );
document.write( "ln(e^m) = m (logarithmic rule)\r<BR>\n" );
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document.write( "ln(y - 1) = ln(x^2) + ln(y - x)<BR>\n" );
document.write( "ln(y - 1) = ln(x^2 * (y - x))<BR>\n" );
document.write( "ln(y - 1) = ln(x^2 * y - x^3)<BR>\n" );
document.write( "you can set y - 1 = yx^2 - x^3<BR>\n" );
document.write( "y - yx^2 = 1 - x^3<BR>\n" );
document.write( "y * (1 - x^2) = 1 - x^3<BR>\n" );
document.write( "y = (1 - x^3)/(1 - x^2) this does not appear to be the book answer you gave earlier,...\r<BR>\n" );
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document.write( "... so also evaluating ln(y - 1) = 2ln(x) + ln(y) - x<BR>\n" );
document.write( "ln(y - 1) = ln(x^2) + ln(y) - x<BR>\n" );
document.write( "ln(y - 1) = ln(yx^2) - ln(e^x)<BR>\n" );
document.write( "ln(y - 1) = ln[(yx^2)/(e^x)]<BR>\n" );
document.write( "you can set y - 1 = (yx^2)/(e^x)<BR>\n" );
document.write( "(y - 1)/(yx^2) = 1/e^x = e^(-x)<BR>\n" );
document.write( "y/(yx^2) - 1/(yx^2) = e^(-x)<BR>\n" );
document.write( "y/y - 1/y = x^2 * e^(-x)<BR>\n" );
document.write( "1 - 1/y = x^2 * e^(-x)<BR>\n" );
document.write( "-1/y = x^2 * e^(-x) - 1<BR>\n" );
document.write( "1/y = 1 - x^2 * e^(-x)<BR>\n" );
document.write( "y = 1/[1 - x^2 * e^(-x)]<BR>\n" );
document.write( "oh that gives that book answer you gave earlier\r<BR>\n" );
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document.write( " </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );