TITLE: Ambiguity with Logs and Inequalities
QUESTION [5 upvotes]: 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.

REPLY [3 votes]: 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 $b < 1$.