Changing base of an inequality

Question

I was considering,

$$-1 \leq \cos\left(\theta - \frac{\pi}{4} \right) \leq1 $$

Can I do this?

$$2^{-1}\leq 2^{\cos\left(\theta - \frac{\pi}{4} \right)} \leq 2^1 $$

And further, Can I multiply with $\frac{1}{\sqrt{2}}$ in the powers of this inequality, that is

$$2^{\frac{-1}{\sqrt{2}}}\leq 2^{\frac{\cos\left(\theta - \frac{\pi}{4} \right)}{\sqrt{2}}} \leq 2^{\frac{1}{\sqrt{2}}} $$

I found that the above result is correct.

But, I am wondering, Am I allowed to do this in any case, like changing bases to "ANYTHING" and multiplying with "ANYTHING" in the powers of an inequality?

Answer 1

Not quite anything. One of the reasons why this works is because the map $x\mapsto2^x$ is increasing and therefore $a\leqslant b\implies2^a\leqslant2^b$. Furthermore, if $\lambda,c>0$,$$a\leqslant b\implies\lambda a\leqslant\lambda b\implies c^{\lambda a}\leqslant c^{\lambda b}.$$

Answer 2

Note: $$1) \quad a^x<a^y,a>0 \iff \begin{cases}x<y,a>1\\ x>y,0<a<1\end{cases} \\ 2) \quad ax<ay \iff \begin{cases}x<y,a>0\\ x>y,a<0\end{cases}.$$