​ An Algebraic Problem

$1000^x=0.1$

$(10^3)^x=0.1$

$10^(3x)=0.1$

$10^(3x)=1/10$

$10^(3x)=10^-1$

$3x=-1$

$x=-1/3$

1/10*1/1000 = 1/10000

x=1/10000

Solve for x over the real numbers:
1000^x = 0.1

0.1 = 1/10:
1000^x = 1/10

Take the logarithm base 1000 of both sides:
Answer: | x = -1/3