1. WECHS – 13 December 2010

2.  Given that log 2.72 = 0.4346, approximate the following without a calculator: log 0.272, log 272, and log 0.00272.  How would you solve this?

3.  Given that log 2.72 = 0.4346, approximate the following without a calculator: log 0.272, log 272, and log 0.00272.  Use the Product Property:

4.  Given that log 2.72 = 0.4346, approximate the following without a calculator: log 0.272, log 272, and log 0.00272.  Use the Product Property:

5.  Given that log 2.72 = 0.4346, approximate the following without a calculator: log 0.272, log 272, and log 0.00272.  Use the Product Property:

6.  Given that log 2.72 = 0.4346, approximate the following without a calculator: log 0.272, log 272, and log 0.00272.  Use the Product Property:

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

8.  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

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

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

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

12. ◦ So,

13.  EOC practice problems 1-4 at: http://www.ncpublicschools.org/docs/accountability /testing/eoc/sampleitems/alg2/20071207alg2g1. 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.