Is this a proof of "$\log(a^x) = x\log(a)$"?

Question

I'm not sure if it's right, but could this be proof of "$\log(a^x) = x\log(a)$"?

For example $\log(1000) = \log(10*100)=\log(10) + \log(100) = \log(10) + \log(10*10) = \log(10) + \log(10) + \log(10) = 3\log(10)$

Answer 1

But $$\log((-2)^2)=2\log|-2|$$ so $$\log(x^2)=2\log|x|$$ and $$x\ne 0$$

Answer 2

To prove that refer to the definition with $a>0$

$$b=\log a^x\iff e^b=a^x \iff e^{b/x}=a \iff \frac b x= \log a \iff b=x\log a$$

Answer 3

If the property $\log pq=\log p+\log q$ is taken for granted, if $x$ is restricted to be a natural and if you make it clear that the computation generalizes to other $a$ and other $x$, then yes, this is an acceptable proof. (Though probably not everybody on this site will agree.)

Answer 4

This is not a proof, your statement is an example. In order to prove a relationship like \begin{equation} \tag{1} \label{1} \log(a^x) = x\log(a) \end{equation}

you have to take into account all possible values of $x$ and $a$ for which \ref{1} should hold. A simple version of the proof you are looking for can be done by using your idea.

Let $x \in \mathbb{N}, a \in (0,\infty)$. By applying the log-identity $\log(a\cdot b) = \log(a) + \log(b)$ repeatedly we obtain: \begin{align} \log(a^x) = \log(a \cdot \dots \cdot a) = \log(a) + \log(a \cdot \dots \cdot a) = x\log(a) \end{align}

An example can suffice as a proof, if it is in the form of a counterexample. Consider the following statement. $$0 < \log(a) \quad \text{for all } a \in [1,\infty)$$

I can prove this statement false by providing the counterexample $a = 1$.

Answer 5

Although an example can often demonstrate the point, it does not count as a proper proof. You can demonstrate why this idea holds using the fact that $\log_b x = y \iff b^y = x$. $$\log_a b^x = y \implies a^y = b^x \implies a^{\frac{y}{x} } = b \implies \log_a b = \frac{y}{x} \implies y = x\log_a b$$ $$\log_a b^x = x\log_a b; b \in \mathbb{R^+}$$