solve exponential equation. express in exact form.
e^x-42e^-x= -19
multiply each term by e^x
(e^x)^2 - 42 + 19e^x = 0
let y = e^x
y^2 + 19y - 42 = 0
(y+21)(y-2) = 0
y = -21 or y = 2
so e^x = 2
x = ln 2
or
e^x = -21, which has no solution
x = ln2
thanks again
To solve the exponential equation e^x - 42e^(-x) = -19, we can use a substitution to simplify the equation.
Let's substitute e^x with a new variable, let's say u. So, we have:
u - 42/u = -19
Now, let's multiply both sides of the equation by u to eliminate the fraction:
u^2 - 42 = -19u
Rearranging the equation, we get:
u^2 + 19u - 42 = 0
Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. In this case, let's factorize it:
(u + 21)(u - 2) = 0
Setting each factor to zero, we have two possible values for u:
u + 21 = 0, which gives u = -21
u - 2 = 0, which gives u = 2
Now, we need to substitute back u with e^x in order to find the solutions for x.
For u = -21:
e^x = -21
However, e^x cannot be negative, so there is no solution for this case.
For u = 2:
e^x = 2
To solve for x, take the natural logarithm (ln) of both sides of the equation:
ln(e^x) = ln(2)
Simplifying, we have:
x = ln(2)
Therefore, the solution to the exponential equation e^x - 42e^(-x) = -19, expressed in exact form, is x = ln(2).