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Solve $$x^2+ \ln^2(x)=1$$

Graph of the given equation enter image description here

From the graph or simply putting $x=1$ satisfy the equation one of the solution is $1$ . How I can find the other solution of the given equation $?$

Abhishek Kumar
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3 Answers3

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First of all, this isn't a quadratic. Secondly, this is a type of an equation that cannot be solved explicitly, that is, to find a closed form expression for $x$. The only way to work such equations around is using a Numerical Analysis approach with a preffered method.

Apart from that, indeed one can see that $x=1$ trully satisfies the equation $x^2 + \ln^2 x - 1 = 0$. One could study the behavior of the function $f(x)$ yielded by that expression, as to know where the other solution is around.

Let $f(x) = x^2 + \ln^2 x - 1$. Then, $f'(x) = 2x - \frac{2\ln x}{x}$. Now, investigate :

$$f'(x) = 0 \Rightarrow 2x - \frac{2 \ln x}{x} = 0 \Rightarrow 2x^2 - 2\ln x =0$$

We've again come across a similar, non-explicitly solvable equation. Working now, as :

$$f''(x) = 0 \Rightarrow 4x - \frac{2}{x} = 0 \Rightarrow 4x^2 - 2 = 0 \Leftrightarrow x = \frac{\sqrt{2}}{2}$$

Note that we excluded the negative part of the solution, as $x >0$ since we are working with a logarithm expression. That, now, can give you a hint on where to look for the other solution and thus how to manipulate/start your preffered numerical method.

Rebellos
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  • Why this is not a quadratic $$x^2+ \ln^2(x)=1$$, I think it is quadratic bcz it can be written like this $$x^2+ 2\ln(x)-1=0$$ and it is quadratic in $x$ – Abhishek Kumar Oct 15 '19 at 07:31
  • Wrong!!! 2 is the power of ln x not x so it can't transform into 2$\ln x$ – Who am I Oct 15 '19 at 07:41
  • @AbhishekKumar Quoting Wikipedia : "In mathematics, the term quadratic describes something that pertains to squares, to the operation of squaring, to terms of the second degree, or equations or formulas that involve such terms." But, when talking about equations/functions, the term quadratic is most oftenly used for the expressions of the kind $ax^2 + bx + c$. This is not such a case, as for starters, the variable $x$ is also involved in a logarithmic expression. By the way $x^2 + \ln^2 x - 1 = 0 \not \Leftrightarrow x^2 + 2\ln x - 1 = 0$. – Rebellos Oct 15 '19 at 13:13
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First of all, $x=1$ is a precise answer. Since this equation can't be solved analytically, instead I considered solving $$x_{n+1}=e^{-\sqrt{1-x_n^2}}$$recursively using Python and by having $x_0=0.1$ I obtained $$x\approx 0.4$$

Mostafa Ayaz
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Well, it's simple

Take $\ln x =k$

So $e^k=x$

Now plugin these equations in ur equation

You'll get $e^{2k}+k^2-1=0$

Now use maclaurin's expansion for $e^{2k}$

So the equation becomes

$$2x+3x^2+\frac{8x^3}{3!} + \frac{16x^4}{4!}+\frac{32x^5}{5!}+\frac{64x^6}{6!}+\frac{128x^7}{7!}=0$$

Used upto degree 7 to get more precise answer

Solve this equation you'll get $k=0,-0.918$

Now $x=e^0=1$ and $x=e^{-0.918}=3.993 \approx 0.4$

Who am I
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