Prove that $\log_{10}(1-x^{-m}) \geq -2 \cdot x^{-m}$ for $x>2^{1/m}$. 
Prove that $\log_{10}(1-x^{-m}) \geq -2 \cdot x^{-m}$ where $m,x > 2^{1/m}$.

I was thinking about the analogous property for the base $e$ logarithm for which it is true that $\log(1-x) >-2x$ for $x<1/2$. But how would we prove this statement?


PRIMER: ELEMENTARY INEQUALITY
In THIS ANSWER, I showed using only the limit definition of the exponential function along with Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le \log(x)\le x}\tag 1$$
for $x>0$. 


For $x>0$ and $m>0$, the domain of the function $f(x)=\log_{10}(1-x^{-m})$ is $x>1$.  If in addition, we have $x>2^{1/m}$, then we can write
$$\begin{align}
\log_{10}(1-x^{-m})&=\frac{\log_e (1-x^{-m})}{\log_e(10)} \tag 2\\\\
&\ge \frac{-x^{-m}}{\log_e(10) (1-x^{-m})} \tag 3\\\\
&\ge \frac{-x^{-m}}{\log_e(10) (1-\left(2^{1/m}\right)^{-m})}\\\\
&=\frac{-2x^{-m}}{\log_e(10)}\\\\
&\ge -2x^{-m}
\end{align}$$
as was to be shown!

In going from $(2)$ to $(3)$, we made use of the left-hand side inequality in $(1)$.

