Solve $x+a^xThe answer to $x+3^x<4$ is $x<1$ by plotting the graph of $y=x+3^x$ and $y =4$. Is there a way to get to the solution algebraically? 
Updated:  is there a way to get to the solution of $x+a^x<b$ in general, algebraically?
 A: The function $x + 3^x$ is strictly increasing and $1 + 3^1 = 4$. (Not sure if this is what you mean by algrebraic method). 
A: The Lambert $W$ Function is defined to be the inverse of $f(x) = xe^x$.  It is not an elementary function, so it cannot be expressed in terms of elementary functions (like $\exp$, $\sin$, $\sqrt[n]{x}$, $x^n$, etc.).  I don't know how exactly you are defining 'algebraic', but whatever your definition is, the Lambert $W$ function is not algebraic.
Let's see if we can solve your problem in terms of the Lambert $W$ function.  Since the Lambert $W$ function is not elementary (algebraic), this will show that we can't solve your equation algebraically.
Let's consider the equation
$$x + a^x = b$$
I'll focus on the case $a \ge 1$, so $x + a^x$ is strictly increasing and the unique solution to this equation is the right endpoint to the infinite interval of points that satisfy $x + a^x < b$.  To use the $W$ function, we'll have to get all of the $x$'s on one side, so we have something like $f(x)e^{f(x)} = C$.
$$e^{x\ln a} = b - x$$
$$e^{b\ln a} = (b-x)e^{(b-x)\ln a}$$
$$a^b\ln a = \ln a(b-x)e^{(b-x)\ln a}$$
Now, using the definition of the $W$ function, we have
$$\ln a(b-x) = W\left(a^b\ln a\right)$$
which we can simplify to get
$$x = b - \frac{W\left(a^b\ln a\right)}{\ln a}$$
For $0 < a < 1$, the situation is more complicated, since 
$$\lim_{x\to+\infty} x + a^x = +\infty$$
and
$$\lim_{x\to-\infty} x+a^x = +\infty$$
so the set of points that satisfy $x + a^x < b$ (if any such points exist) will be bounded above and below.  Since the second derivative of $f(x) = x + a^x$ is $f''(x) = a^x(\ln a)^2$, which is positive everywhere, we can see that the set of solutions will either be a finite open interval $(a,b)$, or the empty set (no solutions).
A: What happens if x=1? Then x+3^x becomes 1+3=4 so that choice is to large because we want less then 4, so anything less than 1 will give us something less than 4. It makes sense !
