spatularificw2

2022-09-08

Why is $\mathrm{ln}({x}^{x})=x\mathrm{ln}(x)$ valid?

I know that $\mathrm{ln}({x}^{k})=k\mathrm{ln}(x)$ for any constant $k$, but why is $\mathrm{ln}({x}^{x})=x\mathrm{ln}(x)$. The exponent $x$ is not constant.

I know that $\mathrm{ln}({x}^{k})=k\mathrm{ln}(x)$ for any constant $k$, but why is $\mathrm{ln}({x}^{x})=x\mathrm{ln}(x)$. The exponent $x$ is not constant.

Collin Gilbert

Beginner2022-09-09Added 11 answers

As x is probably not an integer, ${x}^{x}$ is defined as :

$${x}^{x}={e}^{x\mathrm{ln}(x)}$$

Hence, taking the logarithm give you $\mathrm{ln}{x}^{x}=x\mathrm{ln}(x)$

$${x}^{x}={e}^{x\mathrm{ln}(x)}$$

Hence, taking the logarithm give you $\mathrm{ln}{x}^{x}=x\mathrm{ln}(x)$

Janiah Parks

Beginner2022-09-10Added 1 answers

another way to think about it, for positive real $x,y$

$$\begin{array}{}\text{(1)}& \mathrm{ln}y={\mathrm{log}}_{x}y\cdot \mathrm{ln}x\end{array}$$

and, again by definition

$${\mathrm{log}}_{x}{x}^{x}=x$$

$$\begin{array}{}\text{(1)}& \mathrm{ln}y={\mathrm{log}}_{x}y\cdot \mathrm{ln}x\end{array}$$

and, again by definition

$${\mathrm{log}}_{x}{x}^{x}=x$$

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