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