# Logarithmic Inequality — How do I solve this: Log base 0. 3 (x^2+8) > log base 0. 3 (9x)?

Subham asked the following question involving a logarithmic inequality:

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

SubhamAnd here is our answer:

# A Brush-Up on Small-Base Logarithmic Function

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

# Proof

### Preliminary fact about Small-Base Exponential Function

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$, $x_1 > x_2 \iff b^{x_1} < b^{x_2}$.

### Going Back to the Proof

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 2nd part is proved similarly. $\blacksquare$

# Going Back to the Original Logarithmic Inequality

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

or

which means that .

Pretty neat. Isn’t it? 🙂