Here is an equation that I found is quite impossible to solve without graphing or approximating the answer. $$\sqrt{x} = 1+\ln(5+x)$$ I tried squaring both sides and factoring the $ln$ out, but it was no use. I also tried to get the procedure on WolframAlpha, but it doesn't have a procedure available for me.

  • $\begingroup$ Pretty sure approximations are required. $\endgroup$ Nov 7, 2013 at 3:44
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    $\begingroup$ @zibadawatimmy - So you mean there's absolutely no way to get the exact answer algebraically? $\endgroup$ Nov 7, 2013 at 3:54
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    $\begingroup$ Absolutely? I'm not sure I can go that far, simply because I cannot think of a proof of such a statement. However, the basic problem is that the system is using a transcendental function like $\ln$ in a complicated fashion, which in some sense means the equation is inherently "non-algebraic". $\endgroup$ Nov 7, 2013 at 4:00
  • $\begingroup$ @zibadawatimmy very nicely said. $\endgroup$
    – Betty Mock
    Nov 7, 2013 at 6:16
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    $\begingroup$ It is clear neither $x = 1$ nor $x = -5$ is a solution of $$\sqrt{x} = 1 + \log(5+x) \quad\iff\quad 1\cdot e^{\sqrt{x}-1} - (5+x) \cdot e^0 = 0 $$ For other algebraic $x$ differs from $1$ and $-5$, $\sqrt{x}-1$ and $0$ are distinct algebraic numbers and $5+x$ are non-zero algebraic number. By Lindemann-Weierstrass Theorem, the expression in LHS of $2^{nd}$ expression $\ne 0$. i.e. The equation at hand doesn't have any algebraic solution at all. $\endgroup$ Nov 7, 2013 at 6:19

2 Answers 2


As far as I know, there cannot be any solution to this equation because of the simultaneous presence of $\sqrt{x}$ and $x$. The only solution can be obtained using a numerical method such as Newton.

$$x_{new} = x_{old} - \frac{f(x_{old})}{f'(x_{old})} $$

Starting at $x=1$, which we know is very far from the solution (see the graph of the function), the successive iterates are $6.357, 14.608, 16.536, 16.579$. Starting at $x=10$, the successive iterates are $15.968, 16.575, 16.579.$
If the equation were $x^k=1+\log(5+x^k)$, there is one or more analytical solutions in terms of Lambert function.

  • $\begingroup$ I thought there would be an algebraic way to solve this, but from your answer it seems that approximating by something like Newton's method is the only way to do it, unfortunately. Thanks for the response though. $\endgroup$ Nov 7, 2013 at 5:30
  • $\begingroup$ @Derek朕會功夫. Glad to have been of a little help. $\endgroup$ Nov 7, 2013 at 10:16

Your equation cannot be solved algebraically (= symbolically = in closed form). Wikipedia: Closed form




The left-hand side of the second equation is an algebraic function that depends on $\sqrt(x)$ and $\ln(5+x)$. $\sqrt(x)$ is algebraic for all $x\in\mathbb{C}$, and $\ln(5+x)$ is algebraic only for $x=-4$ (because the value of $\ln(z)$ is algebraic only for $z=1$). But $-4$ is not a solution of your equation. ($-4$ is a solution of the equation $\sqrt{x}-1-\ln(5+x)=-1+2i$.)


$\sqrt(x)$ and $\ln(5+x)$ are algebraically independent over $\mathbb{C}$. Moreover, it is not possible to represent the left-hand side of the second equation as an algebraic function of only one elementary standard function of the variable $x$. Wikipedia: Elementary function

J. F. Ritt proved in 1925 that such a function cannot have an inverse that is an elementary function. Therefore it is not possible to invert the function on the left-hand side by applying only algebraic functions (or algebraic operations) and/or further elementary functions. Therefore it is not possible to solve this equation in this way.


Unfortunately, there is no further commonly used Special function like LambertW known for inverting the left-hand side.


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