Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How do you solve $10^x = x$? I'm not sure how to solve this algebraically. Using log functions wasn't enough.

share|cite|improve this question
    
@JKnecht I don't think most people can try very much and actually get anywhere with this kind of problem. – Simple Art Mar 18 at 18:23
3  
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" – Simple Art Mar 18 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. – JKnecht Mar 18 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. – Simple Art Mar 18 at 18:46

no real solutions

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.

share|cite|improve this answer
1  
Both sides of the coin. (+1) – Simple Art Mar 19 at 23:20

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

share|cite|improve this answer
1  
I'd start with this as well. If you actually get any interception, then continue down the RabbitHole :) – Septronic Mar 20 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 $\boxed{\text{no solution}}.$

share|cite|improve this answer
    
Nice, I guess you could use derivatives if you wanted to show that it grows faster. – Airdish Mar 19 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

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

share|cite|improve this answer
2  
You could use W$_k$ to show each solution separately. – Simple Art Mar 18 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:

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

share|cite|improve this answer

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.

share|cite|improve this answer
    
This question was tagged as algebra-precalculus, so I'm not quiet sure whether the OP has any familiarity with derivatives. – Workaholic Mar 18 at 16:51

from the given equation we get $$x\ln(10)=\ln(x)$$ this is equivalent to $$\ln(10)=\frac{\ln(x)}{x}$$ with this you will find a solution by drawing $$f(x)=\frac{\ln(x)}{x}$$ for $$x>0$$

share|cite|improve this answer
3  
Perhaps I'm missing something, but graphing $\frac{\ln x}{x}$ seems far less trivial than graphing the left- and right-hand sides of the original equation (which are functions the OP should know immediately, in the same way that one would expect the OP to know the multiplication table). However, I suppose one also has to use the fact that the slope of $\ln x$ is decreasing and that it passes through the points $(1,0)$ and $(e,1)$ (and knowing that $(e,1)$ lies below $(1,1)$.) – Dave L. Renfro Mar 18 at 18:00
    
$\ldots$ lies above $(1,1)).$ (I didn't get that last part correctly edited before time ran out on my comment, a feature I think is stupid by the way because it increases the number of errors that get posted in comments.) – Dave L. Renfro Mar 18 at 18:08
    
@DaveL.Renfro Copy-paste the posted comment with the mistake, correct, re-post and delete the initial one. You will have to do mathjax again. – Alecos Papadopoulos Mar 19 at 15:34
    
@Alecos Papadopoulos: I've actually done this 3 or 4 times, and should have thought about it here, but for some reason (that I've forgotten now) I was involved with something else at the time and needed to get back to it, so I just posted my rant and left. And I don't see much point in doing so now, plus others might find it amusing . . . – Dave L. Renfro Mar 21 at 14:46
    
sorry, i don't understand the -1 here – Dr. Sonnhard Graubner Mar 21 at 14:59

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.