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Author Topic: [Mathematics II] Theory Of Logarithms And Indices (Solutions)  (Read 58720 times)

Offline Mr. Babatunde

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The solution to [Mathematics II] Theory Of Logarithms And Indices  via our computer base test.

1. Given that log2 = 0.30103 and log3 = 0.4771, calculate Log 6
   
   Log6 = log (2x3)
   = log2 + log3
   = 0.30103 + 0.4771
   Ans = 0.77815

2. Given that log2 = 0.30103 and log3 = 0.4771, calculate Log 0.9

   Log0.9 = log (9/10)
   = log9 - log10
   = log3^2 - 1
   = 2 log 3 - 1
   = 2 x 0.4771 - 1
   Ans = -0.04576

3. Simplify 2^6 x 2^-6 (Note ^ means Raise to power)

   2^6 x 2^-6 = 2^6 + (-6)
      = 2^6 - 6
      = 2^0
      = 1

4. Simplify 9^1/6 x 9^1/3 (Note ^ means Raise to power)

   9^1/6 x 9^1/3 = 9^1/6 + 1/3
      fins the L.c.m
      = 9^1/2
      = √9
      Ans = 3

5. Simplify (81/16)^-3/4 (Note ^ means Raise to power)

   (81/16)^-3/4 = (16/81)^3/4
   = (4√16/81)^3
   = (2/3)^3
   Ans = 8/27

6. Simplify (2d^3)^2 (Note ^ means Raise to power)

   (2d^3)^2 = 2^2 x (d^3)^2
   = 4d^6

7. Simplify 2a^3 a^4 (Note ^ means Raise to power)

   2a^3 a^4 = 2(a^3 a^4)
   = 2a^(3-4)
   = 2a^-1
   Ans = 2/a

8. Simplify 2/5 Log 32

   2/5 Log 32 = Log(32)^2/5
      = Log (2^5)^2/5
      = Log (2)^2
      Ans = Log 4

9. Simplify 2 - Log 4

   2 - Log 4 = Log (100/4)
      Ans = Log 25

10. Evaluate Log base3 (27)
 
    Log base3 (27) = x
    then 3^x = 27

    3^x = 3^3
    Ans x = 3




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