document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1076450>Question 1076450</A>: <I><SPAN STYLE=\"word-break: break-all;\"> 5^(2x+1)÷25=125<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 #691077 by </B><A HREF=/tutors/aboutme.mpl?userid=addingup><B>addingup(3677)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=addingup><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=addingup><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=691077\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> 5^(2x+1)÷25=125<BR>\n" );
document.write( "multiply both sides times 25<BR>\n" );
document.write( "5^(2x+1) = 3125<BR>\n" );
document.write( "log(5^(2x+1)) = log(3125)<BR>\n" );
document.write( "Remember the rules of logs? The log of a number raised to a power is the power times the log of the number, like this:<BR>\n" );
document.write( "(2x+1)log(5) = log(3125)<BR>\n" );
document.write( "2x+1 = (log(3125))/log(5)<BR>\n" );
document.write( "2x+1 = log_5(3125)<BR>\n" );
document.write( "2x = (log_5(3125))-1<BR>\n" );
document.write( "2x = 5-1<BR>\n" );
document.write( "2x = 4<BR>\n" );
document.write( "x = 2 </SPAN>\n" );
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