Is Wolfram Alpha calculating this incorrectly?

Question

I entered in "does $2ln(x)$ equal $ln(x^2)$" into Wolfram and it came out false. Purplemath.com says that $log_b(m^n) = n · log_b(m)$. Which is correct? And why is there a difference?

Answer 1

For positive real $x$, $$2\ln(x) = \ln(x^2).$$

As others have pointed out, this does not hold for all reals.

The domain of $f(x)=\ln(x^2)$ is $(-\infty, 0)\cup (0,\infty)$, while the domain of $g(x)=2\ln(x)$ is $(0,\infty)$. Functions that have different domains are different functions.

Answer 2

Let $x=-1$. Then, $$2\ln(-1) \Rightarrow \;\; \text{undef}$$ and $$\ln[(-1)^2] = \ln(1) = 0$$

So not always.

Answer 3

No, it isn't.

When you say $f(x) = g(x)$, where $x$ is not a precise value, what is implied is that you are saying $f \equiv g$; that's, $f$ and $g$ are identical as mappings.

This is clearly false here, because $\ln(\cdot)$ is defined only on $(0, \infty)$, whereas $\ln(\cdot^2)$ is defined on $(-\infty, \infty) \setminus \{0\}$.

In brief, your input is not understood as an equality between two values, but as an identification of two functions.

Answer 4

Over the real numbers $\log(x)$ is defined only for $x > 0$. So for example $\log(x^2)$ is $0$ at $-1$, while $2\log(x)$ is undefined at $-1$.

On the other hand, $\log(x^2) = 2\log(x)$ for $x > 0$, and WolframAlpha does tell you so.