WECHS – 13 December 2010.  Given that log 2.72 = 0.4346, approximate the following without a calculator: log 0.272, log 272, and log 0.00272.  How would.

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Presentation transcript:

WECHS – 13 December 2010

 Given that log 2.72 = , approximate the following without a calculator: log 0.272, log 272, and log  How would you solve this?

 Given that log 2.72 = , approximate the following without a calculator: log 0.272, log 272, and log  Use the Product Property:

 Given that log 2.72 = , approximate the following without a calculator: log 0.272, log 272, and log  Use the Product Property:

 Given that log 2.72 = , approximate the following without a calculator: log 0.272, log 272, and log  Use the Product Property:

 Given that log 2.72 = , approximate the following without a calculator: log 0.272, log 272, and log  Use the Product Property:

 Solve the equation 4=3 x using logs in base 3 and base 10.  How do logs allow you to solve for x?

 Solve the equation 4=3 x using logs in base 3 and base 10.  How do logs allow you to solve for x?  Because the Product Property lets you take an exponent out of the log. PRODUCT PROPERTY

 Solve the equation 4=3 x using logs in base 10 and base 3. ◦ First take the log of both sides: ◦ So,

 Solve the equation 4=3 x using logs in base 10 and base 3. ◦ First take the log of both sides: ◦ So, ◦ Finally,

 Our solution works no matter what base you use for the logarithm. What if we change to base 3?

◦ So,

 EOC practice problems 1-4 at: /testing/eoc/sampleitems/alg2/ alg2g1. pdf Problem 1: pure calculator Problem 2: change from log to exponent Problems 3 & 4: harder problems – use logs to solve equations with x in the exponent.