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

Question

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?

Answer 1

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 $x>0$.

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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 $(1)$.

Answer 2

Let $u=x^{-m}$.

$\log_{10}(1-x^{-m})=\log_{10}e\ln(1-u)>\log_{10}e(-2u)>-2u=-2x^{-m}$