document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=133270>Question 133270</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Solve using logs: 2 to the power of 2x +4 minus 2 to the power of 2x = 120\r<BR>\n" );
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document.write( "Please show steps \r<BR>\n" );
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document.write( "Love you guys! </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #97573 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=97573\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 2^(2x+4)=(2^2x)(2^4)=16(2^2x)\r<BR>\n" );
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document.write( "16(2^2x)-2^2x=15(2^2x)\r<BR>\n" );
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document.write( "15(2^2x)=120 __ dividing by 15 __ 2^2x=8 __ taking log base 2 __ log2(2^2x)=log2(8)\r<BR>\n" );
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document.write( "log2(2^2x)=log2(2^3) __ 2x=3 </SPAN>\n" );
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