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Why does the inequality sign change when applying a logarithm on both sides, with the base less than $1$?

I came across the following math which I solved if 2 ways, $$ \left(\frac{1}{2}\right)^n < \frac{1}{4}\\ n\log\left(\frac{1}{2}\right)< \log\left(\frac{1}{4}\right)\\ -0.301n < -0.602 \\ n > 2 $$ The second method is, $$ \left(\frac{1}{2}\right)^n < \frac{1}{4}\\ n\log_\left(\frac{1}{2}\right) \left(\frac{1}{2}\right)< \log_\left(\frac{1}{2}\right) \left(\frac{1}{4}\right)\\ n < 2 \\ $$ Now I know the first one is the correct answer, but what I don't understand why the second method failed to give the correct inequality. Could someone please explain?

Another general question would be, if instead of values they were variable, meaning instead if $\frac{1}{2}$ it was A, and instead of $\frac{1}{4}$ it was B, how would I attempt to solve it since, with the first method, I wouldn't know if $log\left(A\right)$ was negative or positive.

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2 Answers 2

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In the second approach, the base of the $\log()$ is smaller than $1$, which makes the $\log()$ a decreasing function. That's why the inequality symbol needs to change direction.

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You are actually computing:

$$n \log \frac{1}{2} < \log \frac{1}{4}$$ $$\frac{n \log 1/2}{\log 1/2} \color{red}{>} \frac{\log 1/4}{\log 1/2}$$

since $\log \frac{1}{2}$ is negative, and multiplying / dividing by a negative number flips the sign. Since $f(x) = \log x$ is monotonically increasing ($x > 0$) and $\log 1 = 0$, when the base is less than $1$, $\log b$ is negative and you will need to flip the sign.

By the change of base formula, this is equivalent to:

$$n\log_\left(\frac{1}{2}\right) \left(\frac{1}{2}\right) > \log_\left(\frac{1}{2}\right) \left(\frac{1}{4}\right)$$ $$n > 2$$

as you have already said.

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