How to solve 2×exp(x)+2x-7=0 ?

The given equation #2e^(x)+2x-7=0# can be re-written as

#2e^(x) = 7-2x#

Now take these two as separate functions

#f(x)=2e^(x)# and #g(x) = 7-2x# and plot their graph ; their intersection point will be the solution to the given equation #2e^(x)+2x-7=0#

This is shown below :-

Solve

#2e^x + 2x -7 = 0#

Questions like this are in general hard, and the answer depends if you're in Algebra in high school or deeper into math.

For high school, the best approach is to just try some small numbers and see if they work. (This works for many, many high school math problems, fyi.) There is really only one rational #x# that makes #e^x# rational, #x=0#, which isn't a solution. So guessing isn't going to work here.

If an approximation is good enough, we can graph it, or graph #2e^x# and #7-2x# and see where they meet.

Whatever your level, when faced with a hard one like this, it's usually a good move to ask the available expert, which is Wolfram Alpha .

We see Alpha gave us an approximate answer, pretty close to 1, and even a formula using W(x), which the Lambert Product Log, which isn't usually part of high school math.

There's no answer using regular functions and operations we know about in high school Algebra. That's generally true when we add a term with #x# in an exponent to one where #x# appears as a linear or higher power.

That's the end of the answer for most students. But we can go deeper. The product log is an interesting function. Consider the equation

#k = xe^x#

On the right side is an increasing function of #x#, so it will cross #k# sooner or later. Taking the log doesn't really get us anywhere: #ln k = ln x + x #.

We need something like a log, but not one that's the inverse of #e^x#. It needs to be the inverse of #xe^x#. That's called the Product Log or the Lambert W function, defined as:

#k = xe^x# has real solution #x = W(k)#.

We'll restrict our attention to the reals. It's fun to try to discover #W'#s properties. The fundamental one we're given is

#W(xe^x) = x#

Let's let #x=ye^y# in the following so #W(x)=y#. Now

# W(x) e^{W(x)} = y e^y = x #

That's cool. How about

# e^{W(x)} = e^{y} = frac x y = frac{x}{W(x)} #

Taking logs,

# W(x) = ln x - ln W(X) #

# ln W(x) = ln x - W(x) quad # assuming logs are defined

Now that you see what it's like working with W, see if you can use it to solve the equation, or to check Alpha's solution

# x = 7/2 - W(e^(7/2)) #