Solve the exponential equation
e^x + e^-x = 6
let e^x = a
then we have
a + 1/a = 6
a^2 + 1 - 6a = 0
a^2 - 6a = -1
completing the square, (in this case faster than using the formula)
a^2 - 6a + 9 = -1 + 9
(a-3)^2 = 8
a-3 = ± √8
a = 3± 2√2
so e^x = 3 + √8 OR e^x = 3 - √8
using my calculator
x = ln(3+√8) = 1.7627.. or
x = ln(3-√8) = -1.7627..
To solve the exponential equation e^x + e^-x = 6, we can use a substitution method to simplify the equation.
Let's substitute e^x with a variable, say u. Then, e^-x can be expressed as 1/u.
Our equation becomes:
u + 1/u = 6
To eliminate the fraction, we can multiply through by u, which gives us a quadratic equation:
u^2 + 1 = 6u
Rearranging the equation:
u^2 - 6u + 1 = 0
To solve this quadratic equation, we can use the quadratic formula:
u = (-b ± √(b^2 - 4ac)) / 2a
For our equation:
a = 1, b = -6, c = 1
Now, we can substitute these values into the quadratic formula:
u = (-(-6) ± √((-6)^2 - 4*1*1)) / (2*1)
Simplifying further:
u = (6 ± √(36 - 4))/2
u = (6 ± √32)/2
u = (6 ± 4√2)/2
u = 3 ± 2√2
So, we have two possible values for u: 3 + 2√2 and 3 - 2√2.
Since we initially substituted u with e^x, to solve for x, we need to take the natural logarithm (ln) of both sides of the equation:
Case 1:
u = 3 + 2√2
ln(u) = ln(3 + 2√2)
x = ln(3 + 2√2)
Case 2:
u = 3 - 2√2
ln(u) = ln(3 - 2√2)
x = ln(3 - 2√2)
Therefore, the solutions to the exponential equation e^x + e^-x = 6 are x = ln(3 + 2√2) and x = ln(3 - 2√2).