Question

Please explain?

Why can't `a=1`, for `log_a`?

Answer 1

See explanation below

Explanation:

By definition of logarithm `log_ax=y` it's the same to say `a^y=x`

In case of `a=1` the definition would be `1^y=1=x` and would not be a function because one value of x (1) would have infinite values of y associated

Hope this helps

Answer 2

If we consider the definition of the logarithm:

`y=b^x iff log_b(y)=x`

If we put, `b=1`, then:

`y=1^x iff log_1(y)=x`

And clearly `y=1 AA x in RR`, so that the logarithm function collapses.

Another intuitive approach is to consider the logarithm change of base formula:

`log_b(x) -= log_c(x) / log_c(b)`, where `c` is arbitrary

And again, if we choose `b=1` then we get:

`log_1(x) -= log_c(x) / log_c(1)`

And `log1 =0` (any base), causing a division by zero.