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