Hello,
by your equation you calculate the intercepting points of graphs of an exponential function and a line. I've attached a diagram to show the situation.
I don't know a method to solve this equation algebraically. Instead I would use an iteration(?) (I mean that you have to construct a convergent serie to obtain the solution). I'll demonstrate a method which I know under the name Newton's method:
1. Transform the equation into a function: f(x)=e^x-5x+3
2. Choose a x-value for a start. Here I would say:
3. The next x-value is calculated by:
where f'(x) is the first derivation of f. With your problem it is: f'(x) = e^x-5
So you get:
Now plug in this value for
and you'll get the next x-value which comes closer to the solution:
x_1 = 1.3147...
x_2 = 1.4323...
x_3 = 1.4653...
x_4 = 1.4687...
x_5 = 1.4688...
x_6 = is the same as x_5. That means you substract zero from the previous x-value which is only possible if f(x) = 0. That was intended. So you've found one solution.
Change the x_0-value to 2 and you'll get the second solution: x = 1.74375...
Greetings
EB