document.write( "Question 785014: Given the equations, rewrite each as a quadratic equations in u, where u = e^x.
\n" ); document.write( "Solve for u first then for x\r
\n" ); document.write( "\n" ); document.write( "1) e^2x - e^x =6\r
\n" ); document.write( "\n" ); document.write( "2)e^-2x - 3e^-x = -2 \n" ); document.write( "

Algebra.Com's Answer #477361 by fcabanski(1391) $\"\"$ $\"About$

You can put this solution on YOUR website!
If u = $\"e%5Ex\"$ then $\"u%5E2+=+e%5E%282x%29\"$ - when multiplying a base to a power by the same base to a different power, add the powers. $\"e%5Ex+%2A+e%5Ex+=+e%5E%28x%2Bx%29+=+e%5E%282x%29\"$

\n" ); document.write( " $\"e%5E2x+-+e%5Ex+=6+\"$ means $\"u%5E2+-u=6\"$

\n" ); document.write( " $\"u%5E2+-+u+-+6+=+0\"$

\n" ); document.write( "(u-3)(u+2) = 0

\n" ); document.write( "u = 3 and u = -2

\n" ); document.write( " $\"e%5Ex+=+3\"$ Use the inverse function of e which is ln.

\n" ); document.write( " $\"ln%28e%5Ex%29+=+ln%283%29\"$

\n" ); document.write( "x = ln(3) = approximately 1.0986

\n" ); document.write( "For u=-2, do the same. $\"e%5Ex+=+-2\"$

\n" ); document.write( " $\"ln%28e%5Ex%29+=+ln%28-2%29\"$

\n" ); document.write( "x = ln(-2): ln of a negative number is not a real number. So discard this answer. If you want to find the imaginary number solution, it's $\"ln%282%29+%2B+pi%2Ai\"$ = approximately $\".69315+%2B+pi%2Ai\"$

\n" ); document.write( " $\"e%5E%28-2x%29+-+3e%5E%28-x%29+=+-2\"$ is $\"1%2Fu%5E2+-3%2Fu+=+-2\"$

\n" ); document.write( "Multiply all terms by $\"u%5E2\"$ to get $\"1+-+3u+=+-2u%5E2\"$

\n" ); document.write( " $\"2u%5E2+-3u+%2B+1+=+0\"$

\n" ); document.write( "(2u-1)(u-1) = 0

\n" ); document.write( "2u = 1 so u = 1/2 so $\"e%5Ex+=+1%2F2\"$ and thus x = ln(1/2) = approximately -.693

\n" ); document.write( "u=1 so $\"e%5Ex+=+1\"$ so x = ln(1) = 0
\n" ); document.write( " \n" ); document.write( "

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