document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=508900>Question 508900</A>: <I><SPAN STYLE=\"word-break: break-all;\"> What is x in the following equation?<BR>\n" );
document.write( "125^(7x-2)=(1/625)^(9x-5) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #341169 by </B><A HREF=/tutors/aboutme.mpl?userid=Theo><B>Theo(13342)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Theo><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=Theo><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=341169\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> i think i have an answer.<BR>\n" );
document.write( "not sure if this is the way you're supposed to do it, but it appears to work.<BR>\n" );
document.write( "at least i have an answer that appears to be only 1.5 * 10^-10 different which should well be within any rounding requirement you might have.\r<BR>\n" );
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document.write( "here's what i did:\r<BR>\n" );
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document.write( "125^(7x-2) = (1/625)^(9x-5)\r<BR>\n" );
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document.write( "125 = 5^3<BR>\n" );
document.write( "(1/625) = 1/5^4 = 5^-4\r<BR>\n" );
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document.write( "the equation becomes:\r<BR>\n" );
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document.write( "(5^3)^(7x-2) = (5^-4)^(9x-5)\r<BR>\n" );
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document.write( "by the law of exponents that says that (x^m)^n = x^(m*n), the equation becomes:<BR>\n" );
document.write( "5^(3*(7x-2)) = 5^(-4*(9x-5))<BR>\n" );
document.write( "multiplying the exponents out gets us:<BR>\n" );
document.write( "5^(21x-6) = 5^(-36x+20)<BR>\n" );
document.write( "since the base is the same, this means that the equation is equal if and only if the exponents are equal.<BR>\n" );
document.write( "this means that:<BR>\n" );
document.write( "21x-6 = -36x+20<BR>\n" );
document.write( "add 36x to both sides of this equation and add 6 to both sides of this equation to get:<BR>\n" );
document.write( "21x + 36x = 20 + 6 which simplifies to:<BR>\n" );
document.write( "57x = 26<BR>\n" );
document.write( "divide both sides of this equation by 57 to get:<BR>\n" );
document.write( "x = 26/57\r<BR>\n" );
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document.write( "that's your answer.\r<BR>\n" );
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document.write( "substitute for x in the original equation to get:<BR>\n" );
document.write( "125^(7x-2) = (1/625)^(9x-5) becomes:<BR>\n" );
document.write( "125^(7*26/57-2) = (1/625)^(9*26/57-5)<BR>\n" );
document.write( "this simplifies to:<BR>\n" );
document.write( "125^(68/57) = (1/625)^(-51/57)<BR>\n" );
document.write( "this further simplifies to:<BR>\n" );
document.write( "317.3780282 = 317.2780281<BR>\n" );
document.write( "i stored the numbers internally and then subtracted them from each other.<BR>\n" );
document.write( "this time the difference came out 0.<BR>\n" );
document.write( "the reason is that i stayed with fractions until the end so there was no intermediate rounding error.<BR>\n" );
document.write( "that appears to be your answer.<BR>\n" );
document.write( "the solution was to get the bases to be the same and then apply the exponential law of (x^m)^n = x^(m*n).<BR>\n" );
document.write( "since the equation was one of equality and the base was the same, this means that the exponents had to be equal which is how the problem was solved.\r<BR>\n" );
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document.write( " </SPAN>\n" );
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