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

By mathvault | General Math

Logarithm Function with Base 0.3, Logarithmic Inequality

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:

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:

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


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:

  1. $\log_b (x_1) < \log_b (x_2) \longrightarrow x_1 > x_2$
  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.  $\blacksquare$

Going Back to the Original Logarithmic Inequality

Logarithm Function with Base 0.3, Logarithmic Inequality

Note that the logarithmic function of base 0.3 is strictly decreasing.

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

$x^2 + 8 < 9$


$(x-1)(x-8) < 0$ 

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


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