1. Assignment, red pen, pencil, highlighter, GP notebook Solve for x. 1)2)3) total: +1 +2 +1 +2

2. If b is a positive number other than 1, then _____________ if and only if _______. Property of Equality for Logarithmic Functions: Steps: Example #1: 1. When you have log b ()=log b () (and the bases are equal), set the arguments equal. 2. Solve the new equation for x. –x 5x = x + 12 4x = 12 x = 3 44

3. If b is a positive number other than 1, then _____________ if and only if _______. Property of Equality for Logarithmic Functions: Steps: Example #2: 2. Solve the new equation for x. –x 2x = x + 7 x = 7 Complete Practice #1. 1. When you have log b ()=log b () (and the bases are equal), set the arguments equal.

4. Practice #1: –2y 3y – 5 = 2y + 3 y – 5 = 3 a)b) Solve for the variable. c) +5 y = 8 –2x 3x – 1 = 2x + 3 x – 1 = 3 +1 x = 4 Cannot solve because the bases do not match. FYI: Do not “cross out” the logs. We are not dividing out or subtracting out the logs. We are just using the equality property to solve each problem. This is similar to how we solved exponential problems:

5. Example #3: Solve Extraneous Solutions: When solving problems with logarithms, we must _________ check our answers because not all answers may be solutions. ALWAYS x 2 – 14 = 5x x 2 – 5x – 14 = 0 (x – 7)(x + 2) = 0 x – 7 = 0 x + 2 = 0 x = 7 x = –2 Check: Everything looks like it checks out. What gives?

6. It appeared that we have _____ solutions. However, from the definition of a logarithm, not only must the ______ be positive, but the __________ must also be positive. x = 7 x = –2 Therefore, _______ is not a solution, but why??? two arguments base x = –2 To verify our conclusion, take _________ and set it equal to a variable, y, and rewrite it in exponential form. If y = positive #, then the argument should be positive. If y = 0, then the argument should be 1. If y = negative #, then the argument should be a POSITIVE fraction. However, there is nothing we could substitute for y to make 8 y equal –10. Therefore, expressions of this type are ____________. UNDEFINED I pity the fools that don’t check their answers!

7. Practice #2: Solve for x. Be sure to CHECK your answers. a)b) –x 4x + 10 = x + 1 3x + 10 = 1 x = –3 33 –10 3x = –9 Check: negative argument, so x = –3 is not a solution. NO SOLUTION x 2 – 2 = x x 2 – x – 2 = 0 (x – 2)(x + 1) = 0 x – 2 = 0 x + 1 = 0 x = 2 x = –1 Check: x = 2 x = –1 Only solution: x = 2 Any fools need to be pitied? I pity the fool that thinks there are two solutions to this one! Don’t be a turkey, check your answers!

8. +, –, 0 + + + + b x = y log b y = x Commit the following to memory to help you whenever solutions are checked: The exponent can be any real number (positive, negative, zero) However, the base and argument must always be a positive number.

9. Mixed Practice: 1. Solve for x. Be sure to CHECK your answers a)b) –x 3x – 4 = x + 6 2x – 4 = 6 x = 5 22 +4 2x = 10 Be sure to at least mentally check your solutions! (You don’t have to show the work.) –5x 5x + 4 = –3x 4 = –8x –8

10. No solution Mixed Practice: 1. Solve for x. Be sure to CHECK your answers c)d) +4x 15 – 4x = –x 15 = 3x x = 5 33 –x 2x – 3 = x + 2 x – 3 = 2 +3 x = 5 NO SOLUTION, Sucka! Did you check?

12. Mixed Practice: 1. Solve for x. Be sure to CHECK your answers e)f) –x 4x – 10 = x – 1 3x – 10 = –1 x = 3 33 +10 3x = 9 3x – 1 = 2x 2 0 = (2x – 1)(x – 1) 2x – 1 = 0 x – 1 = 0 2x = 1 x = 1 0 = 2x 2 – 3x + 1

13. Mixed Practice: 1. Solve for x. Be sure to CHECK your answers g)h) x 2 + 36 = 100x 2 – 6 = x (x – 3)(x + 2) = 0 x – 3 = 0 x + 2 = 0 x = 3 x = –2 x 2 – x – 6 = 0 x = 3 –36 x 2 = 64 x = ± 8 Did you check? Only x = 3, Sucka!

14. Mixed Practice: 1. Solve for x. Be sure to CHECK your answers i) x 2 – 6 = 2x + 2 (x – 4)(x + 2) = 0 x – 4 = 0 x + 2 = 0 x = 4 x = –2 x 2 – 2x – 8 = 0 x = 4 Did you check? Only x = 4, Sucka!