blestimd4pz

2022-04-03

Stuck on a log equation
I am having trouble figuring out how I can solve this log:
$6\mathrm{log}\left({x}^{2}+1\right)-x=0$
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:
$6\mathrm{log}\left({x}^{2}+1\right)=x$
step 2: divide by 6:
$\mathrm{log}\left({x}^{2}+1\right)=\frac{x}{6}$
step 3: make both sides an exponent of 10 to get rid of the log:
${x}^{2}+1={10}^{\frac{x}{6}}$
step 4. ??????????????

Ruben Gibson

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

Cody Hart

To find the biggest root, you can also iterate the equation as follows:
${x}_{n+1}=6\mathrm{log}\left({x}_{n}^{2}+1\right)$
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:
${x}_{n+1}=\sqrt{{10}^{\frac{{x}_{n}}{6}}-1}$
Finally, you could show by an analysis of the function $6\mathrm{log}\left({x}^{2}+1\right)-x$ that these are the only three possible solutions.

Do you have a similar question?