document.write( "Question 115137: Solve the following algebraically: e^x-6e^-x=1; this is what I have thus far.\r
\n" ); document.write( "\n" ); document.write( "e^x(e^x-6e^-x)=e^x(1)
\n" ); document.write( "e^2x-6=e^x
\n" ); document.write( "e^2x-e^x-6=0\r
\n" ); document.write( "\n" ); document.write( "then use a quadratic equation to solve:\r
\n" ); document.write( "\n" ); document.write( "-(-1) +-square root (-1)^2-4(1)(-6)/2(1)
\n" ); document.write( "1+-square root 1+24/2
\n" ); document.write( "1+-square root 25/2
\n" ); document.write( "1+-5/2 = 1.0986\r
\n" ); document.write( "\n" ); document.write( "Is this correct? \n" ); document.write( "

Algebra.Com's Answer #83762 by jim_thompson5910(35256) $\"\"$ $\"About$

You can put this solution on YOUR website!
You're on the right track. But remember $\"u=e%5Ex\"$ which means $\"x=ln%28u%29\"$ . Since you cannot take the log of a negative number, the only value that will work is $\"u=%281%2B5%29%2F2=6%2F2=3\"$ \r
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\n" ); document.write( "\n" ); document.write( "So our only answer is $\"x=ln%283%29\"$ which is approximately x=1.0986122886681 \n" ); document.write( "

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