document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=239794>Question 239794</A>: <I><SPAN STYLE=\"word-break: break-all;\"> find the value of x<BR>\n" );
document.write( "1.)logx (3x+10)=2 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #175921 by </B><A HREF=/tutors/aboutme.mpl?userid=JimboP1977><B>JimboP1977(311)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=JimboP1977><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=JimboP1977><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=175921\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> I got a different answer to vleith. In the question I assumed that log has a base of x? Is this incorrect?\r<BR>\n" );
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document.write( "log(base x) (3x+10) = 2   (This means what number power to base x gives 3x+10)\r<BR>\n" );
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document.write( "3x+10 = x^2\r<BR>\n" );
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document.write( "10 = x^2 - 3x <BR>\n" );
document.write( "10= (x-(3/2))^2 - 9/4<BR>\n" );
document.write( "(49/4)^(1/2) = x - (3/2)<BR>\n" );
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document.write( "x = +3.5 + 3/2  OR x = -3.5 + 3/2<BR>\n" );
document.write( "x = 5           OR x = -2\r<BR>\n" );
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document.write( " </SPAN>\n" );
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