document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1132820>Question 1132820</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Solve the equation lg(3^x - 2 ^ (4-x )) = 2 + (1/3)lg 8 - (1/4)xlg16\r<BR>\n" );
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document.write( "This is how far I got \r<BR>\n" );
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document.write( "lg(3^x - 2 ^ (4-x )) = 2 + lg(2^3)^1/3 - lg(2^4)^(1/4)x<BR>\n" );
document.write( "                             = 2 + lg2 - lg2^x\r<BR>\n" );
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document.write( "What do I do next? Thanks </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #749962 by </B><A HREF=/tutors/aboutme.mpl?userid=Alex.33><B>Alex.33(110)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Alex.33><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=Alex.33><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=749962\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Assume lg refers to the logarithm with a base of 10. I'll show the fastest way I can think up, somewhat similar to where you are at.<BR>\n" );
document.write( "lg(3^x-2^(4-x))=2+(1/3)lg8-(1/4)xlg16 <BR>\n" );
document.write( "lg(3^x-2^(4-x))=lg(100)+(1/3)lg(2^3)-(1/4)xlg(2^4)<BR>\n" );
document.write( "lg(3^x-2^(4-x))=lg(100)+lg2-xlg2<BR>\n" );
document.write( "lg(3^x-2^(4-x))=lg(200)-lg(2^x)<BR>\n" );
document.write( "lg(3^x-2^(4-x))=lg(200/(2^x))<BR>\n" );
document.write( "3^x-(2^4/2^x)=200/(2^x)<BR>\n" );
document.write( "6^x-16=200<BR>\n" );
document.write( "6^x=216<BR>\n" );
document.write( "x=3\r<BR>\n" );
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document.write( " </SPAN>\n" );
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