Question

find the value of log2 64+ log2 32-log2 16​

Answer 1

Step-by-step explanation:

$log_{2}(64) + log_{2}(32) - log_{2}(16) \\ = log_{2}( {2}^{6} ) + log_{2}( {2}^{5} ) - log_{2}( {2}^{4} ) \\ = log_{2}( {2}^{6 + 5} ) - log_{2}( {2}^{4} ) \\ = log_{2}( {2}^{11} ) - log_{2}( {2}^{4} ) \\ = 11 log_{2}(2) - 4 log_{2}(2) \\ = 11 \times 1 - 4 \times 1 \\ = 11 - 4 = 7.$