Solution of $\log_{\frac12}(x) > 4$

Question

Find all x for: $\log_{\frac12}(x) > 4$

My solution:

$\log_{\frac12}(x) > \log_{\frac12}(\frac{1}2^4)$

$x > (\frac{1}2^4)$

$x>\frac{1}{16}$

This is my solution but this is wrong. The answer is $x<16$. Can anyone explain me where the signs of the inequality changes? I've entered the question in symbolab which resulted in: https://www.symbolab.com/solver/step-by-step/%5Clog_%7B%5Cfrac%7B1%7D%7B2%7D%7D%5Cleft(x%5Cright)%3E4. I understand their solution but I still don't understand why mine is false.

Answer 1

We have that $\log_a(x)$ is increasing if $a>1$ and decreasing if $0<a<1$. So you should reverse the inequality sign when you are cancelling out the logarithms.

Answer 2

HINT

Recall that

$$\log_{\frac12}(x) =\frac{\log x}{\log \frac12}$$

and more in general

$$\log_{a}b =\frac{\log b}{\log a}$$

indeed by definition

$$c=\log_{a}b\iff a^c=b \iff \log a^c =\log b \iff c\log a = \log b \iff c=\frac{\log b}{\log a}$$

Answer 3

The first issue is that $(1/2)^4$ is not $16$, but $1/16$. The second is that if the base is smaller than $1$, you need to reverse the sign of the inequality:$$\log_{\frac12}x>\log_{\frac12}\frac{1}{16}$$ implies $$x<\frac{1}{16}$$

Answer 4

By definition, $y=\log_{\frac12}x\iff x=\dfrac1{2^y}$.

Now the function $\;\dfrac1{2^y}\;$ is decreasing, so, if $y>4$, then $\;x=\dfrac1{2^y}<\dfrac1{2^{4}}$.