document.write( "Question 689684: Solve the exponential equation algebraically. Show steps\r
\n" ); document.write( "\n" ); document.write( "2e^x=5-e^-x \n" ); document.write( "
Algebra.Com's Answer #426080 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
\"2e%5Ex=5-e%5E%28-x%29\"
\n" ); document.write( "I am going to multiply each side by \"e%5Ex\". You will better understand why if I first rewrite \"e%5E%28-x%29\" as \"1%2Fe%5Ex\":
\n" ); document.write( "\"2e%5Ex=5-1%2Fe%5Ex\"
\n" ); document.write( "Multiplying:
\n" ); document.write( "\"e%5Ex%2A%282e%5Ex%29=e%5Ex%2A%285-1%2Fe%5Ex%29\"
\n" ); document.write( "we get:
\n" ); document.write( "\"2e%5E%282x%29=5e%5Ex+-1\"
\n" ); document.write( "Since the exponent of \"e%5E%282x%29\" is exactly twice the exponent of \"e%5Ex\" this equation is in what is called quadratic form. To see this better I am going to use a temporary variable. Let \"q+=+e%5Ex\". Then \"q%5E2+=+%28e%5Ex%29%5E2+=+e%5E%282x%29\". Substituting these in we get:
\n" ); document.write( "\"2q%5E2+=+5q-1\"
\n" ); document.write( "The equation is obviously quadratic. We first want one side to be zero. Subtracting the entire right side from each side we get:
\n" ); document.write( "\"2q%5E2-5q+%2B1=0\"
\n" ); document.write( "Next we factor or use the Quadratic Formula. This will not factor so we must use the formula:
\n" ); document.write( "\"q+=+%28-%28-5%29+%2B-+sqrt%28%28-5%29%5E2-4%282%29%281%29%29%29%2F2%282%29\"
\n" ); document.write( "which simplifies as follows:
\n" ); document.write( "\"q+=+%28-%28-5%29+%2B-+sqrt%2825-4%282%29%281%29%29%29%2F2%282%29\"
\n" ); document.write( "\"q+=+%28-%28-5%29+%2B-+sqrt%2825-8%29%29%2F2%282%29\"
\n" ); document.write( "\"q+=+%28-%28-5%29+%2B-+sqrt%2817%29%29%2F2%282%29\"
\n" ); document.write( "\"q+=+%285+%2B-+sqrt%2817%29%29%2F4\"
\n" ); document.write( "which is short for:
\n" ); document.write( "\"q+=+%285+%2B+sqrt%2817%29%29%2F4\" or \"q+=+%285+-+sqrt%2817%29%29%2F4\"
\n" ); document.write( "We now have solutions for q. But we are looking for solutions for x. Now we substitute back in for q. (It was temporary, remember?)
\n" ); document.write( "\"e%5Ex+=+%285+%2B+sqrt%2817%29%29%2F4\" or \"e%5Ex+=+%285+-+sqrt%2817%29%29%2F4\"
\n" ); document.write( "Now we solve these equations for x. (NOTE: Since \"sqrt%2817%29\" is less than 5, the right side of both equations will be positive. If we had gotten zero or a negative for the right side of either equation there would be no solution for that equation since \"e%5Ex\" can never be zero or negative.)

\n" ); document.write( "We solve for x by finding the ln of each side:
\n" ); document.write( "\"ln%28e%5Ex%29+=+ln%28%285+%2B+sqrt%2817%29%29%2F4%29\" or \"ln%28e%5Ex%29+=+ln%28%285+-+sqrt%2817%29%29%2F4%29\"
\n" ); document.write( "On the left sides we use a property of logs, \"log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29\", to move the exponent out in front:
\n" ); document.write( "\"x%2Aln%28e%29+=+ln%28%285+%2B+sqrt%2817%29%29%2F4%29\" or \"x%2Aln%28e%29+=+ln%28%285+-+sqrt%2817%29%29%2F4%29\"
\n" ); document.write( "And since ln(e) = 1 these become:
\n" ); document.write( "\"x+=+ln%28%285+%2B+sqrt%2817%29%29%2F4%29\" or \"x+=+ln%28%285+-+sqrt%2817%29%29%2F4%29\"
\n" ); document.write( "These are exact expressions for the solutions to your equation. If you want/need decimal approximations get out your calculator.

\n" ); document.write( "P.S. Once you have done a few of these quadratic form equations, you will not need to use a temporary variable. You will see how to go from
\n" ); document.write( "\"2e%5E%282x%29=5e%5Ex+-1\"
\n" ); document.write( "to
\n" ); document.write( "\"2e%5E%282x%29-5e%5Ex+%2B1=0\"
\n" ); document.write( "to
\n" ); document.write( "\"e%5Ex+=+%28-%28-5%29+%2B-+sqrt%28%28-5%29%5E2-4%282%29%281%29%29%29%2F2%282%29\"
\n" ); document.write( "etc. \n" ); document.write( "
\n" );