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It is of course possible to write exponential equations that are difficult for many of you to see, but is it possible to find the exact value? Or can they only be predicted by some method?

If we cannot find its exact value, what methods are available to approximate it?

$$x^x+2x=5$$

$$x^3=2^x$$

$$\text{etc.}$$

Blue
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kirismasdada
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    Please ask only one question per post; here you seem to want at least three answers? – amWhy Dec 25 '22 at 18:44
  • they all have the same difficulty, they are similar – kirismasdada Dec 25 '22 at 18:45
  • I'm especially curious about the first two. – kirismasdada Dec 25 '22 at 18:45
  • They are very different from one another. – amWhy Dec 25 '22 at 18:46
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    Such equations generally don't have solutions in closed form. You can get arbitrarily good approximate solutions by any of a number of techniques – Newton's Method is one of the simplest and one of the best. Every Calculus textbook tells you how to use Newton's Method (and there are pages devoted to it all over the web). – Gerry Myerson Dec 25 '22 at 18:54
  • See also https://math.stackexchange.com/questions/3898662/what-is-this-kind-of-numerical-method-called-and-what-is-the-intuition-behind-it and https://math.stackexchange.com/questions/163721/solve-for-x-2x-x3 and https://math.stackexchange.com/questions/2910250/solving-2x-x3-algebraically – Gerry Myerson Dec 25 '22 at 18:57
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    $x^x=ax+b$ and $e^{x^r}=a x+b$ both have Lagrange reversion expansions – Тyma Gaidash Dec 25 '22 at 18:58
  • Firstly, those are two very different problems. Secondly, most of the time there is no analytical way to solve these kinds of problems nor is there a closed form solution to them. You can get a reasonably close approximation for it using several different methods. One popular method is called Newton's algorithm, as described by Gerry above. – 冥王 Hades Dec 25 '22 at 21:25
  • https://math.stackexchange.com/questions/4586412/how-can-we-show-that-az-ez-and-a-ln-z-z-have-no-elementary-inverse https://math.stackexchange.com/questions/2309691/equations-solvable-by-lambert-function/2527410#2527410 – IV_ Dec 26 '22 at 12:41

1 Answers1

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You can solve the equations you mentioned with special functions such as the Lambert W function aka Omega function $\operatorname{W}\left( x \right)$ and Wright W function aka Wright Omega $\omega\left( x \right)$ function for x, e.g.:

$$ \begin{align*} x^{3} &= 2^{x} ~~~~~\quad\qquad\qquad\mid\quad \cdot 2^{-x}\\ x^{3} \cdot 2^{-x} &= 1\\ x^{3} \cdot e^{- \ln\left( 2 \right) \cdot x} &= 1 ~~~\qquad\qquad\qquad\mid\quad \left( \right)^{\frac{1}{3}}\\ x \cdot e^{-\frac{\ln\left( 2 \right)}{3} \cdot x} &= 1 ~~~\qquad\qquad\qquad\mid\quad \cdot \left( -\frac{\ln\left( 2 \right)}{3} \right)\\ -\frac{\ln\left( 2 \right)}{3} \cdot x \cdot e^{-\frac{\ln\left( 2 \right)}{3} \cdot x} &= -\frac{\ln\left( 2 \right)}{3}\\ \left( -\frac{\ln\left( 2 \right)}{3} \cdot x \right) \cdot e^{-\frac{\ln\left( 2 \right)}{3} \cdot x} &= -\frac{\ln\left( 2 \right)}{3} ~~~~\quad\qquad\mid\quad \operatorname{W}\left( \right)\\ -\frac{\ln\left( 2 \right)}{3} \cdot x &= \operatorname{W}\left( -\frac{\ln\left( 2 \right)}{3} \right) ~~\quad\mid\quad \cdot \left( -\frac{3}{\ln\left( 2 \right)} \right)\\ x &= -\frac{3 \cdot \operatorname{W}\left( -\frac{\ln\left( 2 \right)}{3} \right)}{\ln\left( 2 \right)}\\ x &\approx 1.3735 ~~~~\qquad\qquad\mid\quad \text{via using the main branch}\\ x &\approx 9.9395 ~~~\qquad\qquad\mid\quad \text{via using the branch } -1\\ \end{align*} $$

However, many equations that contain exponential functions also have no (until now found) non-approximate solutions. The easiest way to find these solutions is usually using Newton's method, e.g.:

$$ \begin{align*} x^{x} + 2 \cdot x = 5 &\quad\mid\quad -5\\ x^{x} + 2 \cdot x - 5 = 0 &\quad\mid\quad f\left( x \right) := x^{x} + 2 \cdot x - 5\\ x_{n + 1} = x_{n} - \frac{f\left( x_{n} \right)}{f'\left( x_{n} \right)}\\ x_{n + 1} = x_{n} - \frac{x_{n}^{x_{n}} + 2 \cdot x_{n} - 5}{x_{n}^{x_{n}} \cdot \left( \ln\left( x_{n} \right) + 1 \right) + 2} &\quad\mid\quad \text{chose some } x_{1} \text{ like } x_{1} := 1\\ \lim_{{n} \to{\infty}} x_{n + 1} = \lim_{{n} \to{\infty}} x_{n} - \frac{x_{n}^{x_{n}} + 2 \cdot x_{n} - 5}{x_{n}^{x_{n}} \cdot \left( \ln\left( x_{n} \right) + 1 \right) + 2} \approx 1.5349 &\\ \end{align*} $$

Kevin Dietrich
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