ANSWER
[tex] log_{1000}(32) = 0.502[/tex]
EXPLANATION
We were given that,
[tex] log(32) = 1.505 [/tex]
Recall that the base is 10. So we can rewrite to get,
[tex] log_{10}(32) = 1.505 - - - (1)[/tex]
We now write the logarithm we are trying to evaluate also as logarithm to base 10.
[tex] log_{1000}(32) = log_{ {10}^{3} }(32) [/tex]
Recall that,
[tex] log_{ {p}^{q} }(m) = \frac{1}{q} log_{p}(m) [/tex]
We apply this law to obtain,
[tex] log_{1000}(32) = \frac{1}{3} log_{ 10 }(32) [/tex]
We now substitute equation 1 into this last equation to get,
[tex] log_{1000}(32) = \frac{1}{3} (1.505)[/tex]
This simplifies to,
[tex] log_{1000}(32) = 0.502[/tex]
The correct answer is option D.
