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I am having trouble figuring out how I can solve this log:

$$6 \log(x^2+1)-x=0$$

The steps i've thought to take so far are as follows:

step 1: subtract the right most x to the other side of the equation:

$$6 \log(x^2+1) = x$$

step 2: divide by 6:

$$\log(x^2+1)=\frac{x}{6}$$

step 3: make both sides an exponent of 10 to get rid of the log:

$$x^2+1 = 10^{\frac{x}{6}}$$

step 4. ??????????????

Willie Wong
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Matt
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    Note that $0$ is a solution. – Raskolnikov Mar 26 '11 at 22:21
  • The other solution is about $13.6267$. These are the only two real solutions. – Raskolnikov Mar 26 '11 at 22:29
  • @Raskolnikov: are you sure? 0.1690366 and 45.92765242537 look close to solutions – Henry Mar 26 '11 at 23:08
  • there are three solutions according to my textbook and wolframalpha if you wanna paste the equation over there. I just dont know How to get to the solutions – Matt Mar 26 '11 at 23:10
  • @Henry: Could you have used the natural logarithm instead of the logarithm base 10? @Matt: Should you actually use the natural logarithm instead of logarithm base 10? ;p – Raskolnikov Mar 26 '11 at 23:24
  • I did overlook a solution though, about $0.4161$. – Raskolnikov Mar 26 '11 at 23:28
  • @ Raskolnikov, the problem came as base 10 in the text. How are you finding the solution?? I am still hung up on this step: (x^2)-(10^(x/6))+1=0 – Matt Mar 26 '11 at 23:38
  • I used a pocket calculator. As already explained by Yuval and PEV, you should use some approximation scheme. PEV suggested Newton's method, I suggest iteration based on the form of the equation. – Raskolnikov Mar 26 '11 at 23:45

3 Answers3

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So $$x^2+1 = 10^{x/6}$$ or $$x^2-10^{x/6} +1 = 0$$ So one can use Newton's Method (i.e. choose the initial point to be $x = 0$).

NebulousReveal
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    I.e. there's no analytic solution, although you can find an approximate solution using various methods. Newton's method is a numerical method for finding approximate solutions. – Yuval Filmus Mar 26 '11 at 22:29
  • How would I solve for x after subtracting the 10^(x/6)? – Matt Mar 26 '11 at 23:10
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To find the biggest root, you can also iterate the equation as follows:

$$x_{n+1}=6\log(x_n^2+1)$$

Start for instance with $x_0=1$.

To find the other root besides the trivial $0$, you can iterate the following equation using the same seed as before:

$$x_{n+1}=\sqrt{10^{x_n/6}-1}$$

Finally, you could show by an analysis of the function $6\log(x^2+1)-x$ that these are the only three possible solutions.

EDIT: Extra reference on fixed point iteration.

Raskolnikov
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$$6\ln(x^2+1)-x=0$$ $$6\ln(e^t)-\sqrt{e^t-1}=0$$ for real $t$: $$6t-\sqrt{e^t-1}=0$$ $$e^t=1+36t^2$$ $$\frac{1}{1+36t^2}e^t=1$$

It can be proved that your equation cannot be solved in terms of elementary functions and Lambert W. Moreover, it can be proved that the equation doesn't have solutions except $0$ that are elementary numbers.

The equation can be solved in terms of Generalized Lambert W.

IV_
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