This is my attempt at an algebraic proof of the identity $\frac{1}{x} = \log_{x^x}(x)$ where $x>0$: $$ \text{Let} \space n = \log_{x^x}(x)\space \text{where} \space x>0 $$ $$\log_{x^x}(x)=n$$ $$(x^x)^n=x$$ $$\ln((x^x)^n)=\ln(x)$$ $$xn\cdot \ln(x)=\ln(x)$$ $$xn=1$$ $$n=\frac{1}{x}$$
Asked
Active
Viewed 30 times
0
-
1Hint: Base change formula. $$\log_{x^x}(x) =\frac{\log(x)}{\log (x^x)}$$ – Nothing special Apr 30 '23 at 07:17
-
Your proof is correct, and a "better way" (? at least shorter) is to use What does $ \log_a (b) $ equal to?. More generaly, $\log_{a^n}(a^m) = \frac{m}{n}$. – Anne Bauval Apr 30 '23 at 07:25