- Exponential form: $a^b = c$.
- Logarithmic form: $\log_a(c) = b$.
$1^1 = 1$; therefore, $\log_1(1) = 1$, but it isn't.
Furthermore, isn't the result of $\log_x(x)$ always 1?
Why is there an error? What am I actually doing in $\log_1(1)$?
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3One to the power anything equals one. – velut luna Oct 15 '17 at 21:08
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4$\log_1(x)$ is not defined – windircurse Oct 15 '17 at 21:11
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$\log_1(1)$ could be any number, because any number solves the equation $1^x = 1$. This is a bad thing for a function (which is what we want the logarithm to be), so we leave it undefined.
This is entirely analoguous to $\frac00$ being left undefined, because any number solves $0x = 0$.
Even more analoguously, $\frac a0$ for $a \neq 0$ is being left undefined for the same reason that $\log_1(b)$ for $b \neq 1$ is, although it's a different twain from the one above. This time it's because the corresponding equations $0x = a$ and $1^x = b$ have no solutions.
As to your furthermore, it is rather the case that $\log_\color{red}x(\color{blue}x) = 1$ (as long as $x>0, x\neq 1$). This is because $\color{red}x^1 = \color{blue}x$, and $1$ is the only number that fits in there and still makes it an equality.
Arthur
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