How do I solve $x=1.1^x$? Is it possible? I'm trying to find the intersection of $y=x$ and $y=1.1^x$. I've tried logging both sides and doing every log law on it, but can't figure it out. I know these equations intersect twice thanks to desmos. Using desmos, when I set y=nx, where n is a big number, because the point of intersection is off-screen, I have to scroll a lot. Plus, I'm just curious how to solve this equation, $x=1.1^x$. Also, would the solving process be similar for $x=1.1^{(x-1)}$.
 A: There is a famous function called the Lambert W function which one of the applications deals with solving equations of the slightly more general form $$a^x = kx + m$$
Maybe you can use the derivation in the article to solve your problem?
A: Well... an approximate solution is $x = 38.229\ldots$, which I got by experimentation. 
I knew that experimentation would work because letting
$$
f(x) = x
$$
and
$$
g(x) = 1.1^x
$$
I was able to check that $f(2) > g(2)$, but $f(50) < g(50)$ by direct computation, and because both are continuous, I could tell that there must be a number $a \in [2, 50]$ with $f(a) = g(a)$. 
In fact, there's a really easy way, in a situation like this, to find the value of $a$ to any degree of accuracy you like: you let $h(x) = f(x) - g(x)$, and now you know that $h(2)$ and $h(50)$ have opposite signs. 
You pick the midpoint, $26$, and evaluate $h(26)$. One of three things happens:


*

*$h$ at the midpoint is exactly $0$, and you stop, delighted!

*$h$, at the midpoint, has the opposite sign from $h(2)$

*$h$, at the midpoint, has the opposite sign from $h(50)$
In cases 2 and 3, you replace one of the endpoints with the midpoint so that the signs of $h$ at the two ends are still opposite, and go back to the "pick the midpoint" step and repeat. 
This rapidly converges towards a number $a$ with $h(a) = 0$, so that $f(a) = g(a)$. 
By "rapidly", I mean that in binary notation, you get an extra bit per iteration. That's not blindingly fast, but it's not bad. 
This works because your two functions are continuous, and the Intermediate Value Theorem applies.
If you're hoping for some algebraic trick to let you express the value you're looking for in terms of something "known," my best guess is that someone will tell you it's all about the Lambert $W$ function...whose value you still have to compute numerically.  
A: Your equation can be written in the form
$$x=e^{ax}$$ or
$$a=\frac{\log x}x.$$
The function on the right increases from $-\infty$, peaks at $(e,e^{-1})$ then decreases to $0$.
Hence, for $a<0$, there is a single real solution, smaller than $1$, for $0<a<e^{-1}$ there are two of them (one in $(1,e)$ and the other above $e$), and none for $a>e^{-1}$.
