PRECALCULUS ( EXPONENTIALS & LOGARITHMS )
EXPONENTIALS AND LOGARITHMS[A self-test with answers]1. Evaluate : 2-3 . 30. 250.52. Find the value of 3. Find the value of 4. Evaluate (i) log381 (ii) log7 (iii) log55. Given that log2x = 7 . Find the value of x6. Solve:7. Evaluate (i) (ii) 8. Prove that 9. Show that10. Given , Find the value of A11. Solve:3log2x + log227 = log512512. Solve : 13. solve : 3 log4(x-2) = 614. Given f(x) = 3 . 4x Using laws of Indices explain why 16f(x) = f(x+2)15. Solve : ANSWERSQn 1 : 5/8Qn 2 : 8/27Qn 3 : ¼Qn 4 : (i) 4 (ii) ½(iii) -1Qn 5 : x = 128Qn 6 : x = 1/3Qn 7 : (i) 0 (ii) -2Qn 8 : LHS = 2.3x+1 = 2. (3x . 3) = ( 2 . 3 ).3x = 6 . 3x = RHSQn 9 : Let log4x = k 4k = x x = 22k Therefore, RHS = ½ log2x = ½ log2(22k) = ½ . 2k = k = LHSQn10: 45Qn 11: 2/3Qn 12: x = 1/9Qn 13: x = 18Qn 14: RHS = f(x+2) = 3.4x+2 [replacing x by x+2 ]= 3.4x.16 = 48.4x = 16.(3.4x) = 16f(x) = LHSQn 15: x = -3***************************************If anyone is interested in getting extra problems /online tuition on this topic, you can contact me at :georgeignatiusxx@gmail.com [ G mail]georgeignatius9[Skype]gntsgeorge@yahoo.co.in [ Yahoo]Tuition fee: US $ 8.00 per hour, payable to my Pay Pal Account before each hour taught . It may not take more than 10 hours to cover this topic. *****************************************
Description
This is a self-test for all students from Grade 10 to 12. Answers are provided at the end of this test
A public lesson also is scheduled on this topic on Tuesday, 23rd November, 2010 at 8 am (IST). You are quite free to attend.
***************************************
If anyone is interested in getting extra problems /online tuition on this topic, you can contact me at :
georgeignatiusxx@gmail.com [ G mail]
georgeignatius9 [Skype]
gntsgeorge@yahoo.co.in [ Yahoo]
Tuition fee: US $ 8.00 per hour, payable to my Pay Pal Account before each hour taught . It may not take more than 10 hours to cover this topic.
*****************************************
Presentation Transcript
Your Facebook Friends on WizIQ