document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=969493>Question 969493</A>: <I><SPAN STYLE=\"word-break: break-all;\"> log_2 (x+3)+log_2(x-3)=4<BR>\n" );
document.write( "Solve for x </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #592375 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><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=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=592375\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> x=5\r<BR>\n" );
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document.write( "Start by rewriting as Log (b2) {(x+3)(x-3)}=4.  Adding logs is like multiplying them with the same base.  \r<BR>\n" );
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document.write( "Think of log 10 +log 100=1+2=3;  log 10*100=log 1000=3.\r<BR>\n" );
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document.write( "From the definitions of logs, 2^4=16=(x+3)(x-3)=x^2-9.\r<BR>\n" );
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document.write( "So, x^2-9=16; x^2=25; x+ 5 or -5.  Negative logs don't exist, so -5 is extraneous.\r<BR>\n" );
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document.write( "Try 5 in the original<BR>\n" );
document.write( "Log b2 (8) +log b2 (2)=4 ?<BR>\n" );
document.write( "The first is 3 (2^3=8)  and the second is 1; 3+1=4. </SPAN>\n" );
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