How to properly solve this inequality $2^x < \frac{3}{4}$?

Question

How to properly solve this inequality? $$2^x < \frac{3}{4}$$ I know that it will be something like that: $$ x \stackrel{?}{\ldots} \log_2\frac{3}{4} $$ But I don't know how to decide if it should be $>$ or $<$.

Answer 1

You can apply strictly monotonic increasing functions to both sides of an inequality. This is the case for $\log_2$, so just apply $\log_2$ to both sides.

(A strictly monotonic decreasing function reverses the inequality direction.)

Wikipedia has an article on monotonicity if you want to learn more about it.

Answer 2

hint: $2^x < 2^{\log_{2}\frac{3}{4}} \iff 2^{x-\log_{2}\frac{3}{4}} < 1$

Answer 3

You have $$2^x < \frac{3}{4}$$ equivalently$$\ln 2^x < \ln\frac{3}{4}$$ $$x\ln 2 < \ln\frac{3}{4}$$$$x < \frac{\ln\frac{3}{4}}{\ln 2}$$ $$x < \log_2\frac{3}{4}$$ Observe that $\log_2(\cdot)$ is strictly increasing on $x>0$:$$ (\log_2)'(x)=\frac{1}{\ln 2}\frac{1}x>0, \quad x>0.$$

Answer 4

Generically, $$x^y<a$$

The sinal will be the same if : $$x>1$$

and the sinal will be the opposite if : $$0<x<1$$

Basically, the sinal will stay the same if you have a strictly monotonic increasing function, and will change when you have a strictly monotonic decreasing function.

Answer 5

I'm expanding on what GFauxPas said about monotonicity, notably how you can say whether a function is monotonic and in which direction:

In general, it's a good idea to sketch the curve of the function (in this case $\log_2 x$) to see whether it's increasing or decreasing (if at all). In this case the curve looks like this, from which it is clear that it is strictly increasing on $(0, \infty)$, so we have, $$0 < x < y \implies \log_2 x < \log_2 y$$

Of course, just sketching the curve isn't particularly rigorous, plus this only works if you know the shape of the curve off the top of your head (or if you use a computer). Here are some other ways of justifying this result:

A standard way is using calculus, is to use the fact that if $f'(x) > 0$ for some interval, then the function is strictly increasing on that interval. Since $(\log_2 x)' = 1/(x \ln 2),$ for $x>0$ we can see it is increasing for $x>0$. This method works in general, so it's useful to remember.

Another way is as follows, if $0<x<y,$ then there exists some $n>0$ such that $y = nx.$ Since $x < y = nx,$ it follows that $n>1.$ Then we have, $$ \log_2 y = \log_2 nx = \log_2 n + \log_2 x > \log_2 x$$ Since $\log_2 n > 0.$ Note that this proof does use the fact that $\log_2 n>0$ if $n>1,$ so you may want to think about why that's true.