If log 64 = 1.8061, then the value of log 16 will be (approx)?

Given that, log 64 = 1.8061

 i.e  $\log 43=1.8061$ 

--> 3 log 4 = 1.8061

--> log 4 = 0.6020

--> 2 log 4 = 1.2040

$\Rightarrow \log 42=1.2040$

Therefore, log 16 = 1.2040