Question

Solve the equation `e^x = 3-2x`?

Answer 1

`x = 0.594 \ \ \ (3dp)`

Explanation:

We want to solve:

`e^x = 3-2x => e^x+2x-3 = 0`

Let:

`f(x) = e^x+2x-3`

Our aim is to solve `f(x)=0`. First let us look at the graph:
graph{e^x+2x-3 [-10, 10, -15, 15]}

We can see that there is one in the interval `0 lt x lt 1`. To find the solution numerically, using Newton-Rhapson method we use the following iterative sequence

`{ (x_1,=x_0), ( x_(n+1), = x_n - f(x_n)/(f'(x_n)) ) :}`

Therefore we need the derivative:

`\ \ \ \ \ \ \f(x) = e^x+2x-3`
`:. f'(x) = e^x+2`

So our iterative formula is:

`{ (x_1,=x_0), ( x_(n+1), = x_n - (e^(x_n)+2x_n-3)/(e^(x_n)+2) ) :}`

Then using excel working to 4dp we can tabulate the iterations as follows:

Initial Value `x_0=1`

We could equally use a modern scientific graphing calculator as most new calculators have an "Ans" button that allows the last calculated result to be used as the input of an iterated expression.

And we conclude that the solution is (to 3dp):

`x = 0.594`

Answer 2

See below.

Explanation:

This kind of equation can be explicitly solved, using the Lambert function `W(cdot)`

https://en.wikipedia.org/wiki/Lambert_W_function

Making `y = 3-2x` we have

`e^x=3-2x hArr e^((3-y)/2)=y` or

`e^(3/2)/2 = (y/2) e^(y/2)`

Now

`y = x e^x hArr x = W(y)`

then

`y = W(e^(3/2)/2) = 3-2x` then

`x = 1/2(3-2W(e^(3/2)/2))` or

`x approx 0.5942049585087718`