SOLUTION: (e^x - e^-x)/2 =-2
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Question 372994: (e^x - e^-x)/2 =-2
Answer by jsmallt9(3758) (Show Source): You can put this solution on YOUR website!
One way to solve this is based on recognizing that the left side is sinh(x) which makes your equation:
sinh(x) = -2
(If you have never heard of sinh (hyperbolic sine), then skip down to "Without sinh" below.)
To find x, then we just have to solve
sinh-1(-2) = x
You may have a button for sinh-1 on your calculator. Or you may know the formula:
sinh-1(x) = ln(x + )
which makes
sinh-1(-2) = ln((-2) + )
which simplifies to:
sinh-1(-2) = ln(-2 + )
sinh-1(-2) = ln(-2 + )
which is an exact expression for your answer.
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:
Next we can rewrite the with a positive exponent:
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 :
On the left side we need to use the Distributive Property:
which simplifies to:
Next we add to each side:
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 . 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:
which simplifies as follows:
In long form this is:
or
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 , is a positive number. So we will get a solution from it. We just find the base e logarithm (ln) of each side:
On the left side we use the property of logarithms, , to move the exponent of the argument out in front:
By definition ln(e) = 1 so this simplifies to:
which is the same answer as we got with the hyperbolic sine.
Using a temporary variable
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:
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:
Let q =
Then
Substituting these into our equation we get:
This is obviously a quadratic equation. Using the Quadratic formula on this we get:
which simplifies as follows:
In long form this is:
or
Next we substitute back in for q:
or
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 , is a positive number. So we will get a solution from it. We just find the base e logarithm (ln) of each side:
On the left side we use the property of logarithms, , to move the exponent of the argument out in front:
By definition ln(e) = 1 so this simplifies to:
which is the same answer as we got with both of the solutions above.
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