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?
 A: 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}
A: 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 $a\gt1$.
A: 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.
A: $$\begin{align*}
2^x -2^3 &< 0\\
2^x&<2^3\\
\ln 2^x &< \ln 2^3\\
x\ln 2 &< 3 \ln 2\\
x&<3
\end{align*}
$$
A: This can be simply proven: $2^x - 2^3 < 0 \Leftrightarrow 2^3(2^{x-3} - 1) < 0 \Leftrightarrow 2^{x-3} < 1 \Leftrightarrow x-3 < \log_21 \Leftrightarrow x-3 < 0$
A: 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 ?
