# Inequality sign change with logarithm

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.

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.

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:

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