How to solve the following equation involving an exponential function How do you solve $10^x = x$? I'm not sure how to solve this algebraically. Using log functions wasn't enough.
 A: Hint:
Draw the graph of $y=10^x$ and that of $y=x$, will they ever meet?
A: Before we try to  find numerical solutions, let us first see if there are any solutions in the first place. A quick sketch will show that there should be no solution. Let us prove this algebraically.
First notice that for all real $x,$ $10^{x} > 0.$ A positive is always bigger than a negative, so for $x < 0,$ $10^{x} > x.$
For $x > 0,$ we see that $10^{x}$ grows far faster than $x.$ When $x$ is $0,$ $10^{x} = 1 > 0,$ and beyond that, $10^{x}$ always grows faster. So we have that $10^{x} > x$ for all $x.$ Thus, there is $\boxed{\text{no solution}}.$
A: As others have noted, there are no real solutions in this case.  There are complex solutions.  They are 
$$-{\frac {{\rm W} \left(-\ln  \left( 10 \right) \right)}{\ln  \left( 10
 \right) }}
$$
where $W$ is any branch of the Lambert W function.  The first few solutions in order of increasing real part are 
$$- 0.1191930734 \pm 0.7505832941\,i, 0.5294805081 \pm 3.342716202\,i,
 0.7877834910 \pm 6.083768254\,i, 0.9480581767 \pm 8.821952931\,i,
 1.064691576 \pm 11.55730317\,i$$
A: I can show you some interesting ways to find the answer.
Start with $x=\log_{10}(x)$.
Substitute this into itself to get $x=\log_{10}(\log_{10}(x))$
Repeat this infinitely:  $$x=\log_{10}(\log_{10}(\log_{10}(\dots\log_{10}(x)\dots)))$$
Try plugging in a random number for the $x$ inside the logarithms and use a calculator that can calculate complex numbers with logarithms to find the result.
Plugging in different numbers may produce different answers, but all answers should work to solve $x=10^x$.
For numerical reassurance (I used google)
$$\log(2)=0.30102999566$$
$$\log(\log(2))=-0.52139022765$$
$$\log(\log(\log(2)))=-0.2828+1.3643i$$
$$\log(\log(\log(\dots\log(2)\dots)))=-0.119193073+0.750583294i$$
When I try to plug this back into $x=10^x$, I get that it works, with a very small amount of error.
Also note that:
$$10^x=10^{x\pm\log_{10}(e)2\pi in},n=0,1,2,3,\dots$$
And using that, we get $$x=10^{x+\log_{10}(e)2\pi in}\implies x=\log_{10}(x)\mp\log_{10}(e)2\pi in$$
And putting this into our substitution method:
$$x=\log_{10}(\log_{10}(\dots\log_{10}(x)\mp\log_{10}(e)2\pi in)\mp\log_{10}(e)2\pi in)\mp\log_{10}(e)2\pi in$$
A: 
The graph above should immediately tell you that there are no real solutions to the equation. If you're interested in the complex solutions, here's how we can proceed
$$10^x = x$$
$$1=\frac{x}{10^x}$$
$$1=\frac{x}{e^{\ln 10^x}}$$
$$1=xe^{-x\ln 10}$$
$$-\ln 10=(-x\ln 10)e^{(-x\ln 10)}$$
Therefore
$$W(-\ln 10)=-x\ln 10$$
$$x=-\frac{W(-\ln 10)}{\ln 10}$$
Where $W$ is the Lambert W function.
A: For $x>0$ you have no answers beacause of the derivatives.
For $x<0$ you have no answers beacause $10^x>0>x$.
So there are no answers.
