document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1087279>Question 1087279</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Step-by-step solution of:\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #701535 by </B><A HREF=/tutors/aboutme.mpl?userid=rothauserc><B>rothauserc(4718)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rothauserc><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=rothauserc><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=701535\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 5^(2x+1) = 25^(x) * 5^(3x)<BR>\n" );
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document.write( "Note 25^(x) * 5^(3x) = (5^2)^(x) * 5^(3x) = 5^(2x) * 5^(3x) = 5^(5x)<BR>\n" );
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document.write( "5^(2x+1) = 5^(5x)<BR>\n" );
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document.write( "take logarithms of both sides of =<BR>\n" );
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document.write( "log(5) (2x+1) = 5log(5) (x)<BR>\n" );
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document.write( "divide both sides of = by log(5)<BR>\n" );
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document.write( "2x + 1 = 5x<BR>\n" );
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document.write( "3x = 1<BR>\n" );
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document.write( "x = 1/3 <BR>\n" );
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