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Algebra III Lesson 26 The Logarithmic Form of the Exponential – Logarithmic Equations The Logarithmic Form of the Exponential N = bL N = answer; b = the base; L = power or exponent The exponent of b is L. The power of b is L. The logarithm for b to get N is L. The logarithm function is used to pull down the exponent, not to undo the power. N = bL logb N = L The log base b of N is L Example 26.1 Write 4 = 3y in the logarithmic form. Recall: N = bL So 4 = 3y becomes logb N = L log3 4 = y Example 26.2 Write logx 4 = m in exponential form. N = bL logb N = L So logx 4 = m becomes 4 = xm Logarithmic equations Two key things to watch for when solving these: 1st – Generally make sure the problem is written in exponential form not logarithmic form. 2nd – Many times working both sides of the exponential equation in terms of the same base raised to powers. Example: 16 = 2r 16 = 24 24 = 2r Therefore, r = 4. Example 26.3 Solve: logb 8 = 3 1st rewrite in exponential form N = bL logb N = L So logb 8 = 3 becomes: 8 = b3 This is now a problem we know how to deal with. Take the cube root of both sides. (8 = b3)⅓ 8⅓ = b 2=b Example 26.4 Solve: logb 9 = - ½ Rewrite: 9 = b-½ Square both sides to clear the root. (9 = b-½)2 81 = b-1 Take both sides to the -1 power to clean up the b. (81 = b-1)-1 1/81 = b Example 26.5 Solve: log 3 1 =M 27 Rewrite: 1 = 3M 27 In the previous problems once it was rewritten it liked similar to things we had already done. All that was needed was to take a root. This is different. In this case we need to get both sides in terms of the same number to a power. Since one side is 3M lets try to get the other side as a power of 3. Start with the 27. 27 = 33 So, 1/27 = 1/33 1/33is not quite in terms of 3 to a power. 1/33 = 3-3 This is 3 to a power. So, 1/27 = 1/33 = 3-3 Therefore, if 3-3 = 3M, then M = -3 Example 26.6 Solve: log 1 P = −2 3 Rewrite: −2 1 =P 3 This isn’t to bad to do. −2 1 P = = 32 = 9 3 Example 26.7 Solve: log4 8 = X Rewrite: 8 = 4X Both sides need to becomes some number taken to a power, and it needs to be the same number for both sides. Since, 8 cannot be found by taking 4 to a nice power, what can both sides be converted to as bases? Try powers of 2. 8 = 23 4X = (22)X = 22X So, 8 = 4X becomes 23 = 22X. Which makes 2X = 3, therefore X = 2/3. Practice a) 1] Write kP = 7 in logarithmic form. N = bL logb N = L logk 7 = P 2] Write logk 7 = P in exponential form. kP = 7 b) Solve: logb 27 = 3 Rewrite: b3 = 27 Now cube root both sides. (b3 = 27)⅓ b=3 c) Solve: log 2 1 =M 8 Rewrite: 1 M =2 8 Since the variable is the power, both sides need to be in terms of the same base. And that is going to be 2. Examine the 8, can this be a power of 2? 8 = 23 So, 1 1 = 3 8 2 which becomes 2-3. Therefore, 2M = 2-3, making M = -3 d) Solve: log 1 C = −4 2 Rewrite: −4 1 =C 2 24 = C C = 16 e) Factor: 64x12y6 – 27a6b9 Squares? Are the numbers perfect squares? Are all the powers even? No. Cubes? Are the numbers perfect cubes? Are all the powers multiples of 3? (64 x 12 ) (( ) ( 3 y 6 − 27 a 6b 9 = 4 x 4 y 2 − 3a 2b 3 Use the correct pattern: ( )( )) Yes. 3 a3 - b3 = (a - b) (a2 + ab + b2) = 4 x 4 y 2 − 3a 2b 3 16 x 8 y 4 + 12a 2b 3 x 4 y 2 + 9a 4b 6 )