# Is it correct to move x down in $2^x - 2^3 < 0$?

I have $2^x - 2^3 < 0$ and I think it's correct to conclude that $x - 3 < 0$ but a friend of mind disagree with me. I was wondering if there is such a property or axiom?

• The base-$2$ logarithm is increasing, because of this, you are correct. But it's not 'immediate'. – Git Gud Feb 12 '15 at 16:27
• Since $2^x$ is a strictly increasing function you are right. – Emilio Novati Feb 12 '15 at 16:27
• Generally if $f$ is an increasing function on the interval $I$, then $\forall a,b\in I$, $$f(a)>f(b)\Leftrightarrow a>b$$ – Vim Feb 12 '15 at 16:56
• I like the phrase "friend of mind" :-) – Martin Sleziak Feb 12 '15 at 17:18

Since $\log_2(x)$ is a monotone function, \begin{align} \notag 2^x-2^3 < 0 &\Rightarrow 2^x<2^3\\ \notag &\Rightarrow \log_2(2^x) < \log_2(2^3)\\ \notag &\Rightarrow x<3\\ \notag &\Rightarrow x-3<0. \end{align}

The assertion $$2^x-2^3\lt0\implies x-3\lt0$$

is indeed true. But the general assertion

$$a^x-a^3\lt0\implies x-3\lt0$$

is not. It's only true when $a\gt1$.

Think of it this way. The inequality you have is equivalent to $2^x < 2^3$. Now, the function $f(x)=2^x$ is a strictly increasing function, and therefore $2^x < 2^3 \implies x<3$. So your conclusion is correct, but "move the $x$ down" is not really a proper explanation.

\begin{align*} 2^x -2^3 &< 0\\ 2^x&<2^3\\ \ln 2^x &< \ln 2^3\\ x\ln 2 &< 3 \ln 2\\ x&<3 \end{align*}

This can be simply proven: $2^x - 2^3 < 0 \Leftrightarrow 2^3(2^{x-3} - 1) < 0 \Leftrightarrow 2^{x-3} < 1 \Leftrightarrow x-3 < \log_21 \Leftrightarrow x-3 < 0$

An increasing function is such that $$a<b\iff f(a)<f(b)$$ or if you prefer, $$a-b<0\iff f(a)-f(b)<0.$$

What can you say about "increasingness" of the exponential ?