$x + 3^x < 4$ exponential equation might be trivial for some but I have problems with this equation.
I know the answer for $x + 3^x = 4$ is $x = 1$, however I only see that through visual analysis. I searched for an answer on this specific problem online (since it's in the Michael Spivak Calculus book Chapter 1 Problem 4(xii)) and all I was able to find is "by visual analysis... you can see..." which doesn't answer my question
How can you solve this analytically?
Thanks in advance
 A: From $f'(x)=1+3^x\ln 3>0$, you see that the function is monotonic increasing and cannot have more than one root, which you know to be $x=1$.
Then
$$x<1.$$
A: Consider $f(x) = x+3^x$
$f'(x) = 1 + ln(3)3^x$ on $x \ge 1$
$f'(x) > 0$, so function is growing up. So minimum in $x=1$ (for this set of points).
So for other points function will be less than $4$.
$x<1$
A: In general:
$$x+a^x=b$$
$$-a^x=x-b$$
$$-e^{x\ln a}=x-b$$
Write $b-x=z$:
$$-e^{(b-z)\ln a}=-z$$
$$e^{b\ln a-z\ln a}=z$$
$$e^{b\ln a}=ze^{z\ln a}$$
$$\ln a\cdot e^{b\ln a}=z\ln a\cdot e^{z\ln a}$$
Apply the Lambert W function's defining relation $x=W(x)e^{W(x)}$:
$$z\ln a=W(\ln a\cdot e^{b\ln a})$$
$$z=\frac{W(a^b\ln a)}{\ln a}$$
$$x=b-\frac{W(a^b\ln a)}{\ln a}$$

The above derivation is for $x+a^x=b$. To solve $x+a^x<b$, consider the argument to Lambert's W, $w=a^b\ln a$:


*

*If $w=-\frac1e$ or $w\ge0$ then $W(w)$ can take on only one value. The corresponding solution to $x+a^x=b$ – call it $x_0$ – is such that the inequality holds for $x<x_0$.

*If $-\frac1e<w<0$ then $W(w)$ can take on two values, leading to two solutions for $x+a^x=b$: $x_0$ and $x_1$. The inequality holds for $x_0<x<x_1$.

*Else, $W(w)$ is not defined and the inequality does not hold anywhere.

