Bunuel wrote:

What is the largest integer n such that 3^n < 0.001?

A. −13

B. −9

C. −7

D. 0

E. 3

Given \(3^n<0.001\)

=> Multiplying & dividing RHS by 1000 we have

=> \(3^n<\frac{1}{1000}\)

=> OR \(1000<\frac{1}{3^n}\)

=> OR \(1000<3^{-n}\)

Therefore 'n' should be -ve so that the power of \(3^{-n}\) becomes +ve

Thus

=> if \(n=-6\) then \(3^{-(-6)}=3^6=729\)

=> if \(n=-7\) then \(3^{-(-7)}=3^7=2187\) SUFFICIENT

=> if \(n=-8\) then \(3^{-(-8)}=3^8=6561\)

Since 'n' is -ve largest integer value of 'n' that satisfy \(1000<3^{-n}\) is \(n=-7\)

Option "C"

Thanks

Dinesh