document.write( "Question 372994: (e^x - e^-x)/2 =-2 \n" ); document.write( "
Algebra.Com's Answer #265629 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
\"%28e%5Ex+-+e%5E%28-x%29%29%2F2+=-2\"
\n" ); document.write( "One way to solve this is based on recognizing that the left side is sinh(x) which makes your equation:
\n" ); document.write( "sinh(x) = -2
\n" ); document.write( "(If you have never heard of sinh (hyperbolic sine), then skip down to \"Without sinh\" below.)
\n" ); document.write( "To find x, then we just have to solve
\n" ); document.write( "sinh-1(-2) = x
\n" ); document.write( "You may have a button for sinh-1 on your calculator. Or you may know the formula:
\n" ); document.write( "sinh-1(x) = ln(x + \"sqrt%28x%5E2+%2B+1%29\")
\n" ); document.write( "which makes
\n" ); document.write( "sinh-1(-2) = ln((-2) + \"sqrt%28%28-2%29%5E2+%2B+1%29\")
\n" ); document.write( "which simplifies to:
\n" ); document.write( "sinh-1(-2) = ln(-2 + \"sqrt%284+%2B+1%29\")
\n" ); document.write( "sinh-1(-2) = ln(-2 + \"sqrt%285%29\")
\n" ); document.write( "which is an exact expression for your answer.

\n" ); document.write( "
\n" ); document.write( "\"%28e%5Ex+-+e%5E%28-x%29%29%2F2+=-2\"
\n" ); document.write( "Solving this without recognizing hyperbolic sine or without know how to work with hyperbolic functions, we start by multiplying both sides by 2 (to get rid of the fraction:
\n" ); document.write( "\"e%5Ex+-+e%5E%28-x%29+=-4\"
\n" ); document.write( "Next we can rewrite the \"e%5E%28-x%29\" with a positive exponent:
\n" ); document.write( "\"e%5Ex+-+1%2Fe%5Ex+=-4\"
\n" ); document.write( "By doing this now have another fraction. This gives us a clue about what to do next. We want to eliminate the fraction so we will multiply by \"e%5Ex\":
\n" ); document.write( "\"e%5Ex%28e%5Ex+-+1%2Fe%5Ex%29+=+e%5Ex%28-4%29\"
\n" ); document.write( "On the left side we need to use the Distributive Property:
\n" ); document.write( "\"e%5Ex%2Ae%5Ex+-+e%2Ax%2A%281%2Fe%5Ex%29+=+e%5Ex%28-4%29\"
\n" ); document.write( "which simplifies to:
\n" ); document.write( "\"e%5E%282x%29+-+1+=+-4e%5Ex\"
\n" ); document.write( "Next we add \"4e%5Ex\" to each side:
\n" ); document.write( "\"e%5E%282x%29+%2B4e%5Ex+-+1+=+0\"
\n" ); document.write( "Since the exponent of e in the first term is twice the exponent of e in the second term, this equation is in quadratic form for \"e%5Ex\". This means we can use methods for solving quadratic equations on this equation. (If you have trouble seeing this, see \"Using a temporary variable\" below.) This equation won't factor but we can use the Quadratic Formula:
\n" ); document.write( "\"e%5Ex+=+%28-%284%29+%2B-+sqrt%28%284%29%5E2+-+4%281%29%28-1%29%29%29%2F2%281%29\"
\n" ); document.write( "which simplifies as follows:
\n" ); document.write( "\"e%5Ex+=+%28-%284%29+%2B-+sqrt%2816+-+4%281%29%28-1%29%29%29%2F2%281%29\"
\n" ); document.write( "\"e%5Ex+=+%28-%284%29+%2B-+sqrt%2816+%2B+4%29%29%2F2%281%29\"
\n" ); document.write( "\"e%5Ex+=+%28-%284%29+%2B-+sqrt%2820%29%29%2F2%281%29\"
\n" ); document.write( "\"e%5Ex+=+%28-4+%2B-+sqrt%2820%29%29%2F2\"
\n" ); document.write( "\"e%5Ex+=+%28-4+%2B-+sqrt%284%2A5%29%29%2F2\"
\n" ); document.write( "\"e%5Ex+=+%28-4+%2B-+sqrt%284%29%2Asqrt%285%29%29%2F2\"
\n" ); document.write( "\"e%5Ex+=+%28-4+%2B-+2%2Asqrt%285%29%29%2F2\"
\n" ); document.write( "\"e%5Ex+=+%282%28-2+%2B-+sqrt%285%29%29%29%2F2\"
\n" ); document.write( "\"e%5Ex+=+%28cross%282%29%28-2+%2B-+sqrt%285%29%29%29%2Fcross%282%29\"
\n" ); document.write( "\"e%5Ex+=+-2+%2B-+sqrt%285%29\"
\n" ); document.write( "In long form this is:
\n" ); document.write( "\"e%5Ex+=+-2+%2B+sqrt%285%29\" or \"e%5Ex+=+-2+-+sqrt%285%29\"
\n" ); document.write( "The right side of the second equation is clearly negative. But powers of e cannot be negative. So there are no solutions to the second equation. The right side of the first equation, because \"sqrt%285%29+%3E+2\", is a positive number. So we will get a solution from it. We just find the base e logarithm (ln) of each side:
\n" ); document.write( "\"ln%28e%5Ex%29+=+ln%28-2+%2B+sqrt%285%29%29\"
\n" ); document.write( "On the left side we use the property of logarithms, \"log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29\", to move the exponent of the argument out in front:
\n" ); document.write( "\"x%2Aln%28e%29+=+ln%28-2+%2B+sqrt%285%29%29\"
\n" ); document.write( "By definition ln(e) = 1 so this simplifies to:
\n" ); document.write( "\"x+=+ln%28-2+%2B+sqrt%285%29%29\"
\n" ); document.write( "which is the same answer as we got with the hyperbolic sine.

\n" ); document.write( "Using a temporary variable
\n" ); document.write( "Until you have had some practice, equations in quadratic form can be hard to solve the I did above. A temporary variable can help make what I did easier to understand. With the equation:
\n" ); document.write( "\"e%5E%282x%29+%2B4e%5Ex+-+1+=+0\"
\n" ); document.write( "we still have to recognize that the first exponent is twice the second exponent. But instead of going straight to solving it we can use a temporary variable:
\n" ); document.write( "Let q = \"e%5Ex\"
\n" ); document.write( "Then \"q%5E2+=+%28e%5Ex%29%5E2+=+e%5E2x\"
\n" ); document.write( "Substituting these into our equation we get:
\n" ); document.write( "\"q%5E2+%2B4q+-+1+=+0\"
\n" ); document.write( "This is obviously a quadratic equation. Using the Quadratic formula on this we get:
\n" ); document.write( "\"q+=+%28-%284%29+%2B-+sqrt%28%284%29%5E2+-+4%281%29%28-1%29%29%29%2F2%281%29\"
\n" ); document.write( "which simplifies as follows:
\n" ); document.write( "\"q+=+%28-%284%29+%2B-+sqrt%2816+-+4%281%29%28-1%29%29%29%2F2%281%29\"
\n" ); document.write( "\"q+=+%28-%284%29+%2B-+sqrt%2816+%2B+4%29%29%2F2%281%29\"
\n" ); document.write( "\"q+=+%28-%284%29+%2B-+sqrt%2820%29%29%2F2%281%29\"
\n" ); document.write( "\"q+=+%28-4+%2B-+sqrt%2820%29%29%2F2\"
\n" ); document.write( "\"q+=+%28-4+%2B-+sqrt%284%2A5%29%29%2F2\"
\n" ); document.write( "\"q+=+%28-4+%2B-+sqrt%284%29%2Asqrt%285%29%29%2F2\"
\n" ); document.write( "\"q+=+%28-4+%2B-+2%2Asqrt%285%29%29%2F2\"
\n" ); document.write( "\"q+=+%282%28-2+%2B-+sqrt%285%29%29%29%2F2\"
\n" ); document.write( "\"q+=+%28cross%282%29%28-2+%2B-+sqrt%285%29%29%29%2Fcross%282%29\"
\n" ); document.write( "\"q+=+-2+%2B-+sqrt%285%29\"
\n" ); document.write( "In long form this is:
\n" ); document.write( "\"q+=+-2+%2B+sqrt%285%29\" or \"q+=+-2+-+sqrt%285%29\"
\n" ); document.write( "Next we substitute back in for q:
\n" ); document.write( "\"e%5Ex+=+-2+%2B+sqrt%285%29\" or \"e%5Ex+=+-2+-+sqrt%285%29\"
\n" ); document.write( "The right side of the second equation is clearly negative. But powers of e cannot be negative. So there are no solutions to the second equation. The right side of the first equation, because \"sqrt%285%29+%3E+2\", is a positive number. So we will get a solution from it. We just find the base e logarithm (ln) of each side:
\n" ); document.write( "\"ln%28e%5Ex%29+=+ln%28-2+%2B+sqrt%285%29%29\"
\n" ); document.write( "On the left side we use the property of logarithms, \"log%28a%2C+%28p%5Eq%29%29+=+q%2Alog%28a%2C+%28p%29%29\", to move the exponent of the argument out in front:
\n" ); document.write( "\"x%2Aln%28e%29+=+ln%28-2+%2B+sqrt%285%29%29\"
\n" ); document.write( "By definition ln(e) = 1 so this simplifies to:
\n" ); document.write( "\"x+=+ln%28-2+%2B+sqrt%285%29%29\"
\n" ); document.write( "which is the same answer as we got with both of the solutions above. \n" ); document.write( "
\n" );