# 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.

• @JKnecht I don't think most people can try very much and actually get anywhere with this kind of problem. Commented Mar 18, 2016 at 18:23
• To be honest, I feel as though most people should try searching for the answer first, there are plenty of questions almost titled exactly "How to solve this exponential equation" Commented Mar 18, 2016 at 18:26
• @SimpleArt I think he should have shown where he got stuck when he used the "log functions". Even if his approach was wrong or only a couple of steps. Regarding your other comment i completely agree. Commented Mar 18, 2016 at 18:37
• @JKnecht I'd agree, but when I personally tried to solve this for the first time, the paper I was using went in about 20 different directions, none of which were any closer to the solution than the other. Commented Mar 18, 2016 at 18:46

## 6 Answers

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.

• Both sides of the coin. (+1) Commented Mar 19, 2016 at 23:20

Hint:
Draw the graph of $y=10^x$ and that of $y=x$, will they ever meet?

• I'd start with this as well. If you actually get any interception, then continue down the RabbitHole :) Commented Mar 20, 2016 at 14:33

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 $\boxed{\text{no solution}}.$

• Nice, I guess you could use derivatives if you wanted to show that it grows faster. Commented Mar 19, 2016 at 9:01

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

$$- 0.1191930734 \pm 0.7505832941\,i, 0.5294805081 \pm 3.342716202\,i, 0.7877834910 \pm 6.083768254\,i, 0.9480581767 \pm 8.821952931\,i, 1.064691576 \pm 11.55730317\,i$$

• You could use W$_k$ to show each solution separately. Commented Mar 18, 2016 at 18:24

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:

$$x=\log_{10}(\log_{10}(\dots\log_{10}(x)\mp\log_{10}(e)2\pi in)\mp\log_{10}(e)2\pi in)\mp\log_{10}(e)2\pi in$$

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.

• This question was tagged as algebra-precalculus, so I'm not quiet sure whether the OP has any familiarity with derivatives. Commented Mar 18, 2016 at 16:51