document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=806279>Question 806279</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Can anybody solve log2 x + log2(x &#8722; 2) = 3  those are log base 2\r<BR>\n" );
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document.write( "I tried following instructions and end up with <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2-2x-9 ALIGN=MIDDLE  ALT=\"x%5E2-2x-9\"   DISABLED_style= \"max-width:90%; height:auto;\" > because they say I'm supposed to square the 3, but I don't know if that's right </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #485764 by </B><A HREF=/tutors/aboutme.mpl?userid=htmentor><B>htmentor(1343)</B></A><img src=\"/images/stars/stars-12.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=htmentor><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=htmentor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=485764\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> log(x) + log(x-2) = 3<BR>\n" );
document.write( "Using the product rule we can write the LHS as log of the two factors:<BR>\n" );
document.write( "log(x(x-2)) = 3<BR>\n" );
document.write( "To remove the logarithm, raise the terms on either side of the equation to the 2nd power:<BR>\n" );
document.write( "2^log(x(x-2) = 2^3<BR>\n" );
document.write( "x^2 - 2x = 8<BR>\n" );
document.write( "x^2 - 2x - 8 = 0<BR>\n" );
document.write( "Factor:<BR>\n" );
document.write( "(x-4)(x+2) = 0<BR>\n" );
document.write( "This gives two solutions, x=4 and x=-2<BR>\n" );
document.write( "But since we can't take the logarithm of a negative number, the answer is x=4. </SPAN>\n" );
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