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Exam answers to indices and logarithm question(MTH111)

  1. XLog3base3 = Log3base7
    Recall, Log3 base 3 = 1 (Log of the same base = 1)
    X ร— 1 = Log3base7
    X = Log3base7 ( Find Log 7 base 3 )
    X = 1.78
    (ii) 2^x + 2^xโ€“ยน = 2
    Solution
    2^x + 1/2^x= 2 [ any number raise to power minus sign(-) is inverse of that number]2^x + 1/2^X = 2 ( find l.c.m)
    (4xยฒ + 1) รท2x =2 [ cross multiply ]4ยฒ + 1 =2 ร— 2x
    4xยฒ + 1 = 4x
    4xยฒ โ€“ 4x + 1 =0
    Using factorization method,
    (4xยฒ โ€“ 2x) โ€“( 2x + 1) =0 [ hint: whatโ€™s common in the both side]2x (2x โ€“ 1) -1(2x-1) = 0
    (2x โ€“ 1)(2x-1) = 0
    (2x โ€“ 1) appears twice
    2x โ€“ 1 = 0
    2x = 1 [ divide both side by 2]X = ยฝ(twice)

To prove, put X as ยฝ into the real equation
2^X + 1/2^x = 2
2^ยฝ+ 1/2^ยฝ= 2
1.414 + 0.707 = 2 ( approximately)

b. simplify Log1/5125 (ii) Prove that Logba ร— Logcb ร— Logac = 1
solution

C. If Log2 base5= 0.431 and Log3base 5 =0.682, evaluate Log54/9
Solution
evaluate Log54/9 = Log54 รท Log59
Log522 รท Log532 = 2Log52 รท 2Log53
(2 ร— 0.431) รท ( 2 ร— 0.682)
0.862/1.364 = 0.632( approximately)


Log1/5125 = Log5-1 125


Let the both be in base 5
Log5125 รท Log55-1
Log553 รท Log55-1
3Log55 รท -1Log55
3รท-1 = -3


(ii) Prove that Logba ร— Logcb ร— Logac = 1


Weโ€™ll also apply the same thing to this, only that their bases will be different
Logba ร— Logcb ร— Logac
(Logaa รท Logbb) ร— (Logbb รท Logcc) ร—( Logcc รท Logaa)
As we know Log of the same base = 1
1/1 ร— 1/1 ร— 1/1 = 1 ร— 1 ร— 1
=1

Question on surd

2-1/(โˆš2-1)
Using rationalization method,

2-1/(โˆš2-1) ร— (โˆš2+1)/(โˆš2+1)
2- (โˆš2+1)/ (2+โˆš2-โˆš2-1)
2-(โˆš2+1)/1
2-โˆš2-1
1-โˆš2

YOU CAN DO IT

Oluwamuyide Peter

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