document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=337908>Question 337908</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Ok, I'd like to say thank you in advance. The problem is: log(base 2)(6-x)+ log(base 2)(2-x)=5. I'm not sure if I went about it right but I tried multiplying and got: log(base 2)(x^2 -8x+12)=5. Please help. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #242257 by </B><A HREF=/tutors/aboutme.mpl?userid=scott8148><B>scott8148(6628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=scott8148><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=scott8148><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=242257\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> good so far\r<BR>\n" );
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document.write( "x^2 - 8x + 12 = 2^5 ___ x^2 - 8x + 12 = 32\r<BR>\n" );
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document.write( "x^2 - 8x - 20 = 0\r<BR>\n" );
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document.write( "(x + 2)(x - 10) = 0\r<BR>\n" );
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document.write( "REMEMBER : logarithms are NOT defined for negative quantities<BR>\n" );
document.write( "___ check your answers by plugging back in to the original equation </SPAN>\n" );
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