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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. ??????????????

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    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.2011-03-26

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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 $x = 0$).

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    How would I solve for x after subtracting the 10^(x/6)?2011-03-26
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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.