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

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  • $\begingroup$ @JKnecht I don't think most people can try very much and actually get anywhere with this kind of problem. $\endgroup$ Commented Mar 18, 2016 at 18:23
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    $\begingroup$ 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" $\endgroup$ Commented Mar 18, 2016 at 18:26
  • $\begingroup$ @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. $\endgroup$
    – JKnecht
    Commented Mar 18, 2016 at 18:37
  • $\begingroup$ @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. $\endgroup$ Commented Mar 18, 2016 at 18:46

6 Answers 6

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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.

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    $\begingroup$ Both sides of the coin. (+1) $\endgroup$ Commented Mar 19, 2016 at 23:20
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Hint:
Draw the graph of $y=10^x$ and that of $y=x$, will they ever meet?

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    $\begingroup$ I'd start with this as well. If you actually get any interception, then continue down the RabbitHole :) $\endgroup$
    – Septronic
    Commented Mar 20, 2016 at 14:33
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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 $\boxed{\text{no solution}}.$

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

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

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

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

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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.

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

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