Assignment red pen pencil highlighter GP notebook Solve
Assignment, red pen, pencil, highlighter, GP notebook Solve for x. 1) 2) +1 +1 3) +1 +1 +2 +2 +2 total: +2
Property of Equality for Logarithmic Functions: Steps: If b is a positive number other than 1, then _______ if and only if _______. Example #1: 1. When you have logb()=logb() (and the bases are equal), set the arguments equal. 2. Solve the new equation for x. 5 x = x + 12 –x –x 4 x = 12 4 4 x=3
Property of Equality for Logarithmic Functions: Steps: If b is a positive number other than 1, then _______ if and only if _______. Example #2: 1. When you have logb()=logb() (and the bases are equal), set the arguments equal. 2. Solve the new equation for x. 2 x = x + 7 –x –x x=7 Complete Practice #1.
Practice #1: Solve for the variable. a) b) 3 y – 5 = 2 y + 3 – 2 y 3 x – 1 = 2 x + 3 – 2 x y– 5=3 +5 +5 x– 1=3 +1 +1 y=8 x=4 c) 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:
Extraneous Solutions: When solving problems with ALWAYS check our answers because logarithms, we must _____ not all answers may be solutions. Example #3: Solve x 2 – 14 = 5 x x 2 – 5 x – 14 = 0 (x – 7)(x + 2) = 0 x+2=0 x– 7=0 x = – 2 x=7 Check: Everything looks like it checks out. What gives?
x=7 x = – 2 two solutions. However, from the It appeared that we have _____ base be positive, definition of a logarithm, not only must the ______ arguments must also be positive. but the _____ x = – 2 is not a solution, but why? ? ? Therefore, _______ To verify our conclusion, take _____ and set it equal to a variable, y, and rewrite it in exponential form. I pity the fools that don’t If y = positive #, then the argument should be check their answers! 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. UNDEFINED Therefore, expressions of this type are ______.
Practice #2: Solve for x. Be sure to CHECK your answers. a) b) 4 x + 10 = x + 1 x 2 – 2 = x –x –x 2 – x – 2 = 0 x I pity the fool that thinks 3 x + 10 = 1 (x – 2)(x + 1)solutions =0 there are two – 10 x+1=0 Any fools need to be x – 2 to= this 0 one! 3 x = – 9 pitied? x = – 1 x=2 3 3 x = – 3 Don’t be a turkey, check Check: x = 2 your answers! x = – 1 negative argument, so x = – 3 is not a solution. NO SOLUTION Only solution: x=2
Commit the following to memory to help you whenever solutions are checked: +, –, 0 bx = y + + + logby = x +, –, 0 + The exponent can be any real number (positive, negative, zero) However, the base and argument must always be a positive number.
Mixed Practice: 1. Solve for x. Be sure to CHECK your answers a) b) 3 x – 4 = x + 6 –x –x 2 x – 4 = 6 +4 +4 2 x = 10 2 2 x=5 5 x + 4 = – 3 x – 5 x 4 = – 8 x – 8 Be sure to at least mentally check your solutions! (You don’t have to show the work. )
Mixed Practice: 1. Solve for x. Be sure to CHECK your answers c) d) 15 – 4 x = –x +4 x 15 = 3 x 3 3 x=5 Did you check? No solution NO SOLUTION, Sucka! 2 x – 3 = x + 2 –x –x x– 3=2 +3 +3 x=5
Mixed Practice: 1. Solve for x. Be sure to CHECK your answers e) f) 4 x – 10 = x – 1 –x –x 3 x – 10 = – 1 +10 3 x = 9 3 3 x=3 3 x – 1 = 2 x 2 0 = 2 x 2 – 3 x + 1 0 = (2 x – 1)(x – 1) 2 x – 1 = 0 2 x = 1 x– 1=0 x=1
Mixed Practice: 1. Solve for x. Be sure to CHECK your answers g) h) x 2 – 6 = x x 2 – x – 6 = 0 (x – 3)(x + 2) = 0 x– 3=0 x=3 x+2=0 x = – 2 Did you check? x=3 Only x = 3, Sucka! x 2 + 36 = 100 – 36 x 2 = 64 x=± 8
Mixed Practice: 1. Solve for x. Be sure to CHECK your answers i) x 2 – 6 = 2 x + 2 x 2 – 2 x – 8 = 0 (x – 4)(x + 2) = 0 x– 4=0 x=4 x+2=0 x = – 2 Did you check? x=4 Only x = 4, Sucka!
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