document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=133321>Question 133321</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Solve for x:2 to the power of x^2 -5x = 8\r<BR>\n" );
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document.write( "Please show steps\r<BR>\n" );
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document.write( "Thank you:) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #97511 by </B><A HREF=/tutors/aboutme.mpl?userid=nycsharkman><B>nycsharkman(136)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=nycsharkman><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=97511\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> What you have here is:\r<BR>\n" );
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document.write( "2^(x^2 -5x) = 8\r<BR>\n" );
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document.write( "After going back to the question, I realized that you are talking about an exponential equation.\r<BR>\n" );
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document.write( "This is totally different.\r<BR>\n" );
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document.write( "The first thing you want to do is create the same base on BOTH sides of the equation.\r<BR>\n" );
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document.write( "On the left side we have base 2.\r<BR>\n" );
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document.write( "How can we do that with 8?\r<BR>\n" );
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document.write( "What number multiplied by itself three times will produce 8?\r<BR>\n" );
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document.write( "How about 2?  Yes, 2 to the third power = 8.\r<BR>\n" );
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document.write( "We now have this equation with the same base 2 on both sides:\r<BR>\n" );
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document.write( "2^(x^2 -5x) = 2^3\r<BR>\n" );
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document.write( "Since we now have the same base on both sides of the equation, simply EQUATE the exponents and solve for x.\r<BR>\n" );
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document.write( "x^2 - 5x = 3\r<BR>\n" );
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document.write( "x^2 - 5x - 3 = 0\r<BR>\n" );
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document.write( "This quadratic cannot be factored.  In that case, you can complete the square or use the quadratic formula.  The formula is easier to use.\r<BR>\n" );
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document.write( "x = -b + -sqr{b^2 - 4ac}/2a, where a = 1, b = -5 and c = -3\r<BR>\n" );
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document.write( "x = -(-5)+ - sqrt{(-5)^2 - 4(1)(-3)}/2(1)\r<BR>\n" );
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document.write( "x = 5 + - sqrt{25 + 12}/2\r<BR>\n" );
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document.write( "x = 5 + - sqrt{37}/2\r<BR>\n" );
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document.write( "We now separate x into a positive and negative answer.\r<BR>\n" );
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document.write( "x = 5 + sqrt{37}/2\r<BR>\n" );
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document.write( "OR\r<BR>\n" );
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document.write( "x = 5 - sqrt{37}/2\r<BR>\n" );
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document.write( " </SPAN>\n" );
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