Answer :
To evaluate
[tex]$$
\log\left(10^{-101}\right),
$$[/tex]
we can use a property of logarithms which states that for any positive number [tex]$a$[/tex] and any number [tex]$b$[/tex],
[tex]$$
\log\left(a^b\right) = b \cdot \log(a).
$$[/tex]
Here, [tex]$a = 10$[/tex] and [tex]$b = -101$[/tex]. Since the logarithm is base 10 (as indicated by the notation [tex]$\log$[/tex]),
[tex]$$
\log\left(10^{-101}\right) = -101 \cdot \log(10).
$$[/tex]
We know that
[tex]$$
\log(10) = 1,
$$[/tex]
therefore,
[tex]$$
\log\left(10^{-101}\right) = -101 \cdot 1 = -101.
$$[/tex]
Thus, the answer is [tex]$\boxed{-101}$[/tex], which corresponds to choice (D).
[tex]$$
\log\left(10^{-101}\right),
$$[/tex]
we can use a property of logarithms which states that for any positive number [tex]$a$[/tex] and any number [tex]$b$[/tex],
[tex]$$
\log\left(a^b\right) = b \cdot \log(a).
$$[/tex]
Here, [tex]$a = 10$[/tex] and [tex]$b = -101$[/tex]. Since the logarithm is base 10 (as indicated by the notation [tex]$\log$[/tex]),
[tex]$$
\log\left(10^{-101}\right) = -101 \cdot \log(10).
$$[/tex]
We know that
[tex]$$
\log(10) = 1,
$$[/tex]
therefore,
[tex]$$
\log\left(10^{-101}\right) = -101 \cdot 1 = -101.
$$[/tex]
Thus, the answer is [tex]$\boxed{-101}$[/tex], which corresponds to choice (D).
