document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=407913>Question 407913</A>: <I><SPAN STYLE=\"word-break: break-all;\"> solve the equation (2^x).(2^x+1)=10 correct to 3decimal places<BR>\n" );
document.write( "im getting my answer as 1.161 but it dont tally when i substitute the value for x pliz help </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #287432 by </B><A HREF=/tutors/aboutme.mpl?userid=graphmatics><B>graphmatics(170)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=graphmatics><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=graphmatics><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=287432\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> To solve the equation (2^x).(2^x+1)=10 we first want to try to move the x expression from their exponent positions. The way we will do that is with<BR>\n" );
document.write( "log base 2 (check out Log at http://en.wikipedia.org/wiki/Logarithm).<BR>\n" );
document.write( "We have that<BR>\n" );
document.write( "(2^x).(2^x+1)=10<BR>\n" );
document.write( "LogBase2((2^x)(2^x+1)) = LogBase2(10)<BR>\n" );
document.write( "LogBase2(2^x) + LogBase2(2^x+1) = LogBase2(10)<BR>\n" );
document.write( "xLogBase2(2) + (x+1)LogBase2(2) = LogBase2(10)<BR>\n" );
document.write( "x + x+1 = 3.3219<BR>\n" );
document.write( "2*x = 2.3219<BR>\n" );
document.write( "x = 1.661\r<BR>\n" );
document.write( "\n" );
document.write( "You have the answer is 1.161 but I got 1.661<BR>\n" );
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