document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=115137>Question 115137</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Solve the following algebraically:  e^x-6e^-x=1; this is what I have thus far.\r<BR>\n" );
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document.write( "e^x(e^x-6e^-x)=e^x(1)<BR>\n" );
document.write( "e^2x-6=e^x<BR>\n" );
document.write( "e^2x-e^x-6=0\r<BR>\n" );
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document.write( "then use a quadratic equation to solve:\r<BR>\n" );
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document.write( "-(-1) +-square root (-1)^2-4(1)(-6)/2(1)<BR>\n" );
document.write( "1+-square root 1+24/2<BR>\n" );
document.write( "1+-square root 25/2<BR>\n" );
document.write( "1+-5/2 = 1.0986\r<BR>\n" );
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document.write( "Is this correct? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #83768 by </B><A HREF=/tutors/aboutme.mpl?userid=bucky><B>bucky(2189)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=bucky><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=83768\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Correct, but you made a leap of faith at the very end. You had:<BR>\n" );
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document.write( "1+-5/2 = 1.0986<BR>\n" );
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document.write( "and where did you get that from? The two answers you got from the quadratic equation are:<BR>\n" );
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document.write( "(1+5)/2 = 3 and (1-5)/2 = -4/2 = -2<BR>\n" );
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document.write( "How did you get the 1.0986???? The answer is that you solved for e^x = 3 and e^x = -2. You<BR>\n" );
document.write( "can handle those in the same manner as below. I solved the problem by factoring which isn't <BR>\n" );
document.write( "a whole bunch better than your way, but it is a little different. Maybe you'll find it a<BR>\n" );
document.write( "little easier.<BR>\n" );
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document.write( "Begin factoring at the point where you got:<BR>\n" );
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document.write( "e^2x-e^x-6=0 <BR>\n" );
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document.write( "Think of this as:<BR>\n" );
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document.write( "(e^x)^2 - (e^x) - 6 = 0<BR>\n" );
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document.write( "Factor to get:<BR>\n" );
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document.write( "(e^x -3)(e^x + 2) = 0<BR>\n" );
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document.write( "This will be true if either of the two factors is zero because multiplication by zero on<BR>\n" );
document.write( "the left side makes the left side equal to the right side.<BR>\n" );
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document.write( "So either (e^x - 3) = 0 or (e^x + 2) = 0 will satisfy the equation. Let's do e^x - 3 = 0 first:<BR>\n" );
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document.write( "Add + 3 to both sides to get:<BR>\n" );
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document.write( "e^x = + 3<BR>\n" );
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document.write( "Take the natural log (ln) of both sides:<BR>\n" );
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document.write( "ln(e^x) = ln(3)<BR>\n" );
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document.write( "Bring the exponent out as a multiplier:<BR>\n" );
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document.write( "x*ln(e) = ln(3)<BR>\n" );
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document.write( "Recognize that ln(e) = 1 which makes the equation become:<BR>\n" );
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document.write( "x = ln(3) = 1.098612289<BR>\n" );
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document.write( "This is the same as the answer you got.<BR>\n" );
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document.write( "Next do the other factor:<BR>\n" );
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document.write( "e^x + 2 = 0<BR>\n" );
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document.write( "Subtract 2 from both sides:<BR>\n" );
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document.write( "e^x = -2<BR>\n" );
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document.write( "Take the ln of both sides:<BR>\n" );
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document.write( "ln(e^x) = ln(-2)<BR>\n" );
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document.write( "But you cannot take the ln of a negative number. So ignore this factor.  The only answer<BR>\n" );
document.write( "is x = ln(3). <BR>\n" );
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document.write( "Hope this adds a little info to your knowledge of dealing with the base of the natural logarithms.<BR>\n" );
document.write( "Good work on solving the problem with the quadratic equation. Just fill in the missing<BR>\n" );
document.write( "step to show how you got 1.0986 from the answers you got from the quadratic equation.<BR>\n" );
document.write( ". </SPAN>\n" );
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