# Ambiguity with Logs and Inequalities

Solve the following inequality:

$$0.8^x > 0.4$$

Method 1 (Using the Common Logarithm)

$$\log_{10}0.8^x > \log_{10}0.4$$ $$x\log_{10}0.8 > \log_{10}0.4$$

Because $$\log_{10}0.8 < 0$$

The sign of the inequality will change when we divide both sides of the equation by the LHS of the above inequality:

$$x < \frac{\log_{10}0.4}{\log_{10}0.8}$$

Using a reverse of the change of base theorem, we get:

$$x < \log_{0.8}0.4$$

Method 2 (Using another based Logarithm)

$$\log_{0.8}0.8^x > \log_{0.8}0.4$$ $$x\log_{0.8}0.8 > \log_{0.8}0.4$$

Because $$\log_{0.8}0.8 > 0$$

The sign of the inequality will not change when we divide both sides of the equation by the LHS of the above inequality:

$$x > \frac{\log_{0.8}0.4}{\log_{0.8}0.8}$$

The denominator of the RHS of the inequality is equal to one, therefore:

$$x > \log_{0.8}0.4$$

The sign is reversed using both methods - the textbook states the answer to be the one gotten using the first method, but I can't spot where I went wrong in method two.

Any help will be greatly appreciated, thanks in advance.

• What do you mean by the "reverse" of the change-of-base formula? The change-of-base formula tells us that $\dfrac{\log_{10}0.4}{\log_{10}0.8} \vphantom{\dfrac{\displaystyle\int}{\displaystyle\int}}$ and $\log_{0.8}0.4$ are both the same number. What does the "reverse" tell us? ${}\qquad{}$ – Michael Hardy Aug 14 '15 at 0:11
• You can just use $\log_{0.8}(0.8)=1$ and shorten the second argument significantly. – Double AA Aug 14 '15 at 4:26

The method 2 is wrong at the very beginning because the following is wrong : $$0.8^x\gt 0.4\Rightarrow \log_{0.8}0.8^x\gt \log_{0.8}0.4$$

Instead we have $$0.8^x\gt 0.4\Rightarrow \log_{0.8}0.8^x\color{red}{\lt} \log_{0.8}0.4$$ because $0.8\lt 1$.

For $a\gt 1$, $$b\gt c\gt 0\iff \log_ab\gt \log_ac.$$ For $a\lt 1$, $$b\gt c\gt 0\iff \log_ab\color{red}{\lt} \log_ac.$$

Expanding slightly on what @mathlove says, just plot $\log_{0.8} x$ vs. $x$ (say in Maple or Mathematica if you have them, or in a spreadsheet otherwise). You will see that it is a monotonically decreasing function on $x>0$. Any time we take

$x>y$

and $f(x)$ is monotonically decreasing then

$f(x) < f(y)$

$\log_b (x)$ is monotonically decreasing (on $x>0$) any time $b < 1$.