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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.

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Pretty sure approximations are required. –  zibadawa timmy Nov 7 '13 at 3:44
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@zibadawatimmy - So you mean there's absolutely no way to get the exact answer algebraically? –  Derek 朕會功夫 Nov 7 '13 at 3:54
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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". –  zibadawa timmy Nov 7 '13 at 4:00
    
@zibadawatimmy very nicely said. –  Betty Mock Nov 7 '13 at 6:16
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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. –  achille hui Nov 7 '13 at 6:19

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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.

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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. –  Derek 朕會功夫 Nov 7 '13 at 5:30
    
@Derek朕會功夫. Glad to have been of a little help. –  Claude Leibovici Nov 7 '13 at 10:16

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