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High School Mathematics - 27.2 Laws of Logarithms

 Laws log a(mn) = logam + logan log a(m/n) = logam - logan log amn = nlogam Change of base: logbr = logar/logab Example: Find the value of log216Ö8 Solution: log216Ö8 = log216 + log2Ö8--by first law log216 + log281/2 log216 + 1/2log28-- by third law log224 + 1/2log223 4 + 1/2.3 = 11/2--- because log22 = 1 Answer: 11/2 Directions: Solve the following.

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High School Mathematics - 27.2 Laws of Logarithms

 Q 1: log2Ö6/Ö83/21/21/4 Q 2: log 327Ö729689 Q 3: log11[121Ö14641/3Ö1331]375 Q 4: log103Ö10001/32/52/3 Q 5: Prove that 3log2 + log5 = log 40Answer: Q 6: log23Ö4/42.Ö8-29/61/23/4 Q 7: log54Ö25/625-13/5-7/21/2 Q 8: log73Ö33751 Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only!