# How do I simplify and calculate this inequality? log &#x2061;<!-- ⁡ --> ( x

How do I simplify and calculate this inequality?
$\mathrm{log}\left({x}^{3}\right)>|x-1|$
I can't figure out how to go about solving this inequality, besides this one step:
$3\mathrm{log}\left(x\right)>|x-1|$
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Allison Pena
As said in the comments, you have two situations buit you can notice that for $x=1$, the lhs and rhs are equal.
So consider the case where $x=1-ϵ$ and use Taylor expansion for the lhs at $x=1$. Yous will easily see that the inequality is not satisfied. Since you noticed that for $x=1$, $lhs=rhs$, the inequality will be satisfied for $x>1$. So, you can now forget the absolute value in the rhs.
Now, compute the derivative of $lhs-rhs$; it is equal to $3/x-1$ and so canceled for $x=3$; the second derivative being negative, then this point corresponds to a maximum. For $x=3$, $lhs=3log\left(3\right)$ and $rhs=2$; so, at this point, $lhs>rhs$. On the other side, you know that $x$ moves faster then $log\left(x\right)$; so there is a point which will corresponds to an $x$ intercept. You will then need to solve the equation
$\mathrm{log}\left({x}^{3}\right)=x-1$
$\mathrm{log}\left({x}^{3}\right)=x-1$
which does not any simple analytical solution. If you plot the function, you will see that the solution is close to $x=6.71$
So, the inequality is satisfied for $1