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).