solve the equation algebraically.
7e^x=11-e^-x
7 e^x = 11 - 1/e^x
let e^x = t
7t = 11 - 1/t
times t
7t^2 = 11t - 1
7t^2 - 11t + 1 = 0
t = (11 ± √93)/14
then e^x = (11 + √93)/14 = 1.474546...
x = ln 1.4745.. = .38835..
or
e^x = (11 - √93)/14 = .096882..
x = ln .096882.. = -2.33426...
To solve the equation algebraically, we will try to isolate one of the variables (either "e^x" or "e^-x") on one side of the equation. Let's start by isolating "e^x" on one side.
Step 1: Move all terms involving "e^x" to one side:
7e^x + e^-x = 11.
Step 2: To simplify the equation, let's get rid of the negative exponent. We can achieve this by multiplying both sides of the equation by e^x:
(7e^x)(e^x) + e^-x(e^x) = 11(e^x).
This simplifies to:
7(e^x)^2 + 1 = 11(e^x).
Step 3: Let's substitute a new variable "u" to replace "e^x":
u = e^x.
The equation now becomes:
7u^2 + 1 = 11u.
Step 4: Rearrange the equation to a quadratic form:
7u^2 - 11u + 1 = 0.
Step 5: We have a quadratic equation. We can solve it using factoring, completing the square, or using the quadratic formula. In this case, factoring might not be straightforward, so let's use the quadratic formula:
u = (-b ± √(b^2 - 4ac)) / 2a.
Using our quadratic equation:
u = (11 ± √((-11)^2 - 4(7)(1))) / (2(7)).
Simplifying further:
u = (11 ± √(121 - 28)) / 14.
u = (11 ± √93) / 14.
Step 6: We now have the values of "u." Recall that we set u = e^x. So, we have:
e^x = (11 ± √93) / 14.
To solve for x, we will take the natural logarithm (ln) on both sides:
x = ln[(11 ± √93) / 14].
Therefore, the solution to the equation 7e^x = 11 - e^-x is x = ln[(11 ± √93) / 14].