# Developing computational capacity for students in teaching exponential functions, exponential functions and logarithmic functions in calculus 12 - 15

Activities of teachers

 Student Activities Content Question: Statehow to calculate, solve,... HD7: Similar to HD 5HD 8:Method 1: Solve Method 2: Calculatorx 2  2 x  3Enter 2  8 xThen CALC the input results. Get result AHD 9 : Similar to HD 8 7. Let log a b  3, log a c   2 . Calculate log a xa) x = a 3 b 2 cb) x = a 4 3 bc 3x 2  2 x  38. Find the solution set 2  8 x .A. _ S   1 ;3  . B. S    1;3  . C. S    3;1  . D. S    3  . 9. Find the solution set of the equationl og  x  1   log 1  x  1   ​​1. 22A.   3  13   . B.  3  .   2   C.  2  5;2  5  . D .  2  5  .10. Solution set of inequalities:2019 x  3 x  1  2019 x  2  x 2  3 x  x  3  0 2A.    ;0  B .  C .  3  D.  3 ;   11. Mr. Y saves 75 million in the bank for a term of 3 months and an interest rate of 0.59%/month. If Mr. A does not withdraw interest at all terms, then after 3 years, Mr. A will receive the amount ofA. 92576000 B. 80486000

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Activities of teachers

 Student Activities Content HD 10 : In the solution region, we CALC some values ​​that satisfy answer CH D 11 : S n  A .  1  r  nInterest rate 3.0.59%=1.77%.After 3 years (12 quarters), the amount collected, both principal and interest, is: 92576000 C. 92690000 D. 90930000 Activity 3. Discover, evaluate, explore and solve problems Question: Read HD 12: 12. A person regularly deposits an amount of money T in the bank every month in the form of compound interest at an interest rate of 0.6% per month. Knowing that after 15 months that person will have 10 million VND. Which of the following numbers is the amount T closest to?A. 535,000 B. 635,000C. 613,000 D. 643,000Answer1 0 .0 0 0 .0 0 0  T   1  0, 6 %  1 5  1  .  1  0, 6 %   T  6 3 5 .0 0 0 0 , 6 %  13. Le's mother borrowed 50 million VND from the bank in installments with an interest rate of 1.15%/month for 2 years. subject, division After n months  n  N *  search, search To be : solution. S  A   1  r  n  1   1  r  n r  

Activities of teachers

 Student Activities Content HD 13:The way to calculate the remaining amount after n months is exactly the same as the formula for calculating monthly bank deposits and withdrawals: How much does Ms. Nam have to pay each month?A. 1361313 VND B.136132 VND C. 136131 VND D. 1361312 VND Answer: 1  r  n  1 S  A  1  r  n  X . n r5.10 7 .  1, 0115  48 .0, 0115X   1 , 0115  4 8  1  1361312, 80714. How many positive integer values ​​of the parameter are there for the equationDoes 16 x  2.12 x  ( m  2)9 x  0 have a positive root? A. 1. B . 2 . C. 4 . D. 3 .Answer:Return the equation to:2.9 x  2.12 x  16 x m 9x _- HD: use mode 7- Click MODE 7 to enter into the computer2.9 x  2.12 x  16 x f ( x ) 9x _Select Start -9 end 9 step 1We can see that f(x) is positive in the solution region (0.2; 2.9).

Activities of teachers

 Student Activities Content HD 14:Method 1 : Who solves?- Students: divide into 2 sides.16 x . or 9x- Students solve and find 2 values ​​of m. The result is Method 2: Explore- Teacher: If m = f(x) how do you plan to use the calculator?- Students: can find the hemorrhoidal value range of f(x)- Teacher: What is the most effective way to find the value domain? HD 15:Method 1 : Who solves?- Students divide the two sides4 x or 9 x- Students solve for x and get 2 solutions: x = 0;x  log  2 3  9 2  The result is C Method 2: Explore So m = 1 and m = 2. Answer B.15. Number of solutions of the equation2.4 x  9 x  1  11.6 x A. 0 B. 1 C . 2 D. 3PrizeMethod 2: Enter MODE 7 into the computerf ( x )  2.4 x  9 x  1  11.6 x Select Start -9 end 9 step 1 Get 2 positive f(x) regions, 1 negative f(x) region. Conclusion: There are 2 times f(x) changes sign. The result f(x) = 0 has 2 solutions. Answer C Activity 4. Consolidation

Emphasize:

- Calculate and change formulas: select a, b, c, x, y... arbitrary numbers and use CALC command

- How to solve equations, inequalities and logarithms using the CALC method.

- Apply to general problems by analyzing, evaluating, synthesizing,...

IV. LEARN EXPERIENCE, SUPPLEMENT

+ Students solve equations and inequalities quickly and accurately

+ Students can calculate expressions quickly

+ Students take tests faster.

APPENDIX 5

Grade 12A14 transcript before experimental teaching

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 First and last name Class Code Point first Le Thi Thuy An 12A14 209 5 2 Dinh Nguyen Phuong Anh 12A14 357 7 3 Nguyen Thi Hong An 12A14 209 7 4 Nguyen Thi My Duyen 12A14 132 4 5 Tran Khanh Du 12A14 357 5 6 Nguyen Tien Hai 12A14 209 4 7 Tran Hoang Gia Han 12A14 209 9 8 Tran Tu Huyen 12A14 485 2 9 Phan Thi Cam Huong 12A14 209 5 ten Luc My Linh 12A14 209 5 11 Nguyen Thi Truc Ly 12A14 132 3 twelfth Tran Thi Thuy My 12A14 132 5 13 Phan Ngoc Ngan 12A14 209 5 14 Doan Mong Nghi 12A14 485 6 15 Quach Tieu Nguyet 12A14 132 3 16 La Thi Huyen Nhu 12A14 485 3 17 Nguyen Hoang Phuc 12A14 132 7 18 Tran Cong Son 12A14 132 7 19 Nguyen Tien Tai 12A14 209 7 20 Lam Gia Te 12A14 485 4 21 Nguyen Thanh Tu 12A14 132 4 22 Ho Thanh Tuyen 12A14 485 5 23 Quach Thi Tuyen 12A14 485 6

24

 Trinh Thanh Tuyen 12A14 357 3 25 Ngo Quy Ty 12A14 209 6 26 Nguyen Huu Thai 12A14 132 6 27 Tran Ngoc Thanh 12A14 209 5 28 Du Dong Thanh 12A14 209 5 29 Thi Hong Tham 12A14 132 5 30 Le Minh Thuan 12A14 132 5 thirty first Nguyen Thi Thu Thuy 12A14 485 6 32 Nguyen Anh Thu 12A14 209 5 33 Vǜ Anh Thu 12A14 357 3 34 Vǜ Anh Thu 12A14 485 7 35 Tran Thi Kieu Trang 12A14 132 5 36 Nguyen Thi My Tran 12A14 357 8 37 Phan Minh Trong 12A14 357 5 38 Ngo Thien Trung 12A14 485 3 39 Thai Hoang Vinh 12A14 485 4 40 Tran Duc Vinh 12A14 357 6 41 Bui Thi Yen Vy 12A14 209 7 42 Tran Ngoc Yen Vy 12A14 209 3 43 Mai Ngoc Yen 12A14 132 5 44 HuǶnh Kim Yen 12A14 209 5 45 La Thi Hoang Yen 12A14 209 6

APPENDIX 6

Grade 12A15 transcript before control teaching

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 First and last name Class Code Point first Bui Thi Hoang An 12A15 485 3 2 Nguyen Thi Tuong Duy 12A15 357 5 3 Vo Bao Duy 12A15 485 5 4 Truong Thanh Dong 12A15 357 4 5 Dang Nguyen Hung 12A15 357 5 6 Tran Nguyen Gia Hung 12A15 485 4 7 Tran Quoc Hung 12A15 132 7 8 Duong My Huong 12A15 132 2 9 Le Thi Kieu 12A15 485 7 ten Nguyen Tran Hoang Khang 12A15 357 7 11 To Hoang Khang 12A15 209 3 twelfth Ha Phuoc Khanh 12A15 485 5 13 Mai Truc Linh 12A15 209 5 14 Quach Tuan Loi 12A15 132 6 15 Nguyen Thi Diem My 12A15 209 3 16 Ly Be Ngan 12A15 132 3 17 Danh Xuan Nghi 12A15 132 4 18 Nguyen Ai Ngoc 12A15 357 4 19 Nguyen Thi Bich Ngoc 12A15 209 7 20 Truong Thi Xuan Nguyen 12A15 132 7 21 HuǶnh Thi Dung Nhu 12A15 209 7 22 Le Xuan Binh Phong 12A15 357 5 23 Nguyen Thi Hanh Phuc 12A15 209 6