Write 2log35 + log32 as a single logarithm.

Using the log law a log(z) = log(z^a), 
2 log(35) = log(35^2) = log(1225) 

Using the log law log(a) + log(b) = log(a*b), 
2log35 + log32 = log(1225*32) = log(39200) 

You didn't state what base the logs have, so I assume it's 10. Then the answer in the requested form is log(10)(39200).