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By mathvault | General Math

Subham asked the following question involving a logarithmic inequality:

Solve $\log_{0.3} (x^2+8) > \log_{0.3} (9x)$.

And here is our answer:

When $\log(x)$ has a base that is strictly between $0$ and $1$, it behaves as a strictly decreasing function, meaning that:

$ \displaystyle x_1 > x_2 \iff \log x_1 < \log x_2$

First, notice that while for **large** bases ($>1$) *exponential* functions are strictly increasing, for small bases (i.e., base strictly between $0$ and $1$), exponential functions are actually strictly decreasing. That is:

For all $x_1, x_2 \in \mathbb{R}$ and any small base $ b$, $x_1 > x_2 \iff b^{x_1} < b^{x_2}$.

Given any two *positive* real numbers $x_1, x_2$ and any small base $b$, we break down the proof in two parts:

- $\log_b (x_1) < \log_b (x_2) \longrightarrow x_1 > x_2$
- $\log_b (x_1) \ge \log_b (x_2) \longrightarrow x_1 \le x_2$

For the first part, suppose that $\log (x_1) < \log (x_2)$, then the strictly-decreasing-ness of the exponential function with base $b$ implies that:

$\displaystyle b^{\log_b (x_1)} > b^{\log_b (x_2)}$

Or equivalently,

$ \displaystyle x_1 > x_2 $

So the first part is done. The second part is proved similarly. $\blacksquare$

Applying the strictly-decreasing-ness of $\log_{0.3} (x)$ to the original logarithmic inequality, we have:

$x^2 + 8 < 9$

or

$(x-1)(x-8) < 0$

which means that $1 < x < 8$. Pretty neat. Isn’t it? 🙂

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