Chapter 6 - Vector Spaces - 6.2 Linear Independence, Basis, and Demension - Exercises for 6.2 - Page 457: 13

This set is not linearly independent.

Since $2\ln(2x)-(2\ln(2))\cdot 1 = 2\ln(x)+2\ln(2) - 2\ln(2) = 2\ln(x) = \ln(x^2)$, a vector can be written as a linear combination of other vectors. So, the set of vectors is not linearly independent.