Is the logarithm function $ y = \log_{x} (x) $
always '1' for $x > 0$, this is equal to the implicit function $ x=x^{y} $ so $ y=1$
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what about $\frac{1}{\sqrt{2} }+ \frac{i}{\sqrt{2}}$ – Shullies Apr 16 '22 at 08:50
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You should specify that $x \neq 1$ since $1^x = 1$ does not have a unique solution. – N. F. Taussig Apr 16 '22 at 09:02
1 Answers
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Almost. Due to the logarithm law $$\log_a b=\frac{\log_cb}{\log_c a}$$ just plug in $a=b$.
However, there are two restrictions: $a$, $b$ and $c$ must be positive; and $a$ and $c$ must not be 1 because log to base 1 makes no sense resp. we would divide by 0 in the fraction above.
Hence the answer to your question is "yes, but $x$ must not be 1".
emacs drives me nuts
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