Solve the exponential equation Express you solutions in exact form only Please show your work
e^2x - 9e^x - 20 = 2
(e^x)^2 - 9e^x - 22 = 0
let y = e^x
y^2 - 9x - 22 = 0
(y-11)(y+2) = 0
y = 11 or y = -2
e^x = 11
x = ln 11
or
e^x = -2, no solution ( x = ln(-2) which is undefined)
Thank u!
To solve the exponential equation `e^2x - 9e^x - 20 = 2`, we can use substitution to turn it into a quadratic equation. Let's first set `y = e^x`. So, our equation becomes `y^2 - 9y - 20 = 2`.
Now, we can rearrange the equation by bringing the `2` to the other side: `y^2 - 9y - 22 = 0`.
We now have a quadratic equation in terms of `y`. To solve it, we can either use factoring, completing the square, or the quadratic formula. Factoring might not be immediately obvious, so let's use the quadratic formula.
The quadratic formula states that for an equation `ax^2 + bx + c = 0`, the solutions for `x` are given by `x = (-b ± √(b^2 - 4ac)) / (2a)`.
In our case, `a` is `1`, `b` is `-9`, and `c` is `-22`. Plugging these values into the quadratic formula, we have:
`y = (-(-9) ± √((-9)^2 - 4(1)(-22))) / (2(1))`
Simplifying the equation further, we have:
`y = (9 ± √(81 + 88)) / 2`
`y = (9 ± √169) / 2`
`y = (9 ± 13) / 2`
Solving for `y`, we have two solutions:
1) `y = (9 + 13) / 2 = 22 / 2 = 11`
2) `y = (9 - 13) / 2 = -4 / 2 = -2`
Now that we have the values for `y`, we can substitute back `y = e^x`.
For the first solution, `y = e^x = 11`. Taking the natural logarithm of both sides:
`ln(e^x) = ln(11)`
`x = ln(11)`
So one solution is `x = ln(11)`.
For the second solution, `y = e^x = -2`. However, the exponential function is defined only for positive values, so this solution is not valid. Thus, we discard it.
Therefore, the solution to the equation `e^2x - 9e^x - 20 = 2` is `x = ln(11)`, written in exact form.