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I created an equation a bit ago where I knew the answer, but not how to solve it.

Equation: $$x = 10^{x-1}$$

Answer: $x = 1$

I can not see to find any documentation related to this problem. I know that the equation can be written as $x = a^{x-1}$ and the answer will still be $x = 1$. The only reason I know the answer is because I can plug it in and I made it around the answer. I would love to know the steps behind how to solve this equation.

Jimmy R.
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Gattag
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    Don't think there are easy ways to provide a closed form solution, but there is another root in $(0, 0.5)$, which is evident if you check signs of $10^{x-1}-x$. – Macavity Dec 14 '14 at 17:14
  • you can try the Newton method – Dr. Sonnhard Graubner Dec 14 '14 at 17:24
  • There are no algebraic methods to solve such equations. You can try to plot the functions or make use of numerical methods. There is plenty of information about the latter. – rubik Dec 14 '14 at 17:25

1 Answers1

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You can rewrite your equation as

$$-x\ln 10 e^{-x\ln 10}=-\frac{\ln 10 }{10},$$ and then use Lambert W function to solve the obtained equation: $-x\ln 10 = W\left(-\frac{\ln 10 }{10}\right)$. As you can read on wiki, for some values of $z$ $W(z)$ is multiply defined, hence one one branch you obtain $\forall z>0 W(z\ln z) = \ln z$, which in our case gives $x=1$. Other branches of $W$ will give other roots of this equation.

TZakrevskiy
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