solve the following exponential equation
e^x-4=-e^-x
please show work
if you let u = e^x,
u-4 = -1/u
u^2 - 4u = -1
u^2 - 4u + 1 = 0
now just use the quadratic formula and if you get values for u, wind up with
x = ln(u) if defined
To solve the equation e^x - 4 = -e^(-x), we can follow these steps:
Step 1: Begin by adding e^(-x) to both sides of the equation to isolate e^x on the left side:
e^x - 4 + e^(-x) = 0
Step 2: Now, let's simplify the equation. Since e^(-x) is the reciprocal of e^x, we can multiply the entire equation by e^x to eliminate the denominator:
(e^x)(e^x - 4 + e^(-x)) = 0
Step 3: Distribute e^x into the parentheses:
(e^(2x) - 4e^x + 1) = 0
Step 4: Now, we can rewrite the equation in a quadratic form:
(e^x)^2 - 4(e^x) + 1 = 0
Step 5: Let's substitute y = e^x to simplify the equation:
y^2 - 4y + 1 = 0
Now, we have a standard quadratic equation that can be solved using the quadratic formula or factoring.
Step 6: Using the quadratic formula, which states that for an equation in the form ax^2 + bx + c = 0, the solutions are given by x = (-b ± √(b^2 - 4ac))/(2a), we can obtain the values of y:
y = (-(-4) ± √((-4)^2 - 4(1)(1)))/(2(1))
Simplifying further:
y = (4 ± √(16 - 4))/2
y = (4 ± √(12))/2
y = (4 ± 2√(3))/2
y = 2 ± √(3)
Step 7: Now, we substitute e^x back in for y:
e^x = 2 ± √(3)
To separate the positive and negative solutions, we write two separate equations:
e^x = 2 + √(3) and e^x = 2 - √(3)
Step 8: Take the natural logarithm of each side to solve for x:
ln(e^x) = ln(2 + √(3))
x = ln(2 + √(3))
Similarly, for the second equation:
ln(e^x) = ln(2 - √(3))
x = ln(2 - √(3))
Therefore, the solutions to the exponential equation e^x - 4 = -e^(-x) are x = ln(2 + √(3)) and x = ln(2 - √(3)).