How to solve indices and logarithm equation on MTH 111→ Past question and answer

THE DEPARTMENT OF MATHEMATICS AND STATISTICS
FIRST SEMESTER TEST 2020/2021
CLASS: ND1 civil Engineering
COURSE CODE: MTH 111
COURSE TITLE: Algebra and Element

1. 3^x = 7; XLog3base3 = Log7base3
   Recall, Log3base3 = 1 (reason is that, Log of the same base = 1)
   X × 1 = Log7base3
   X = Log7base3 ( Find Log 7 base 3 ) [check the picture above, to get the clearer view of the question]
   X = 1.78
   (ii) 2^x + 2^-x = 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)
   (4x^2 + 1)/2x =2 [ cross multiply ]
   4x^2 + 1 =2 × 2x
   4x^2 + 1 = 4x
   4x^2 – 4x + 1 =0
   Using factorization method,
   (4x^2 – 2x) –( 2x + 1) =0 [ hint: what’s common in the both side]
   2x (2x – 1) -1(2x-1) = 0
   (2x – 1)(2x-1) = 0
   (2x – 1) appears twice
   2x – 1 = 0
   2x = 1 [ divide both side by 2]
   X = ½(twice)

To check if our answer is correct, put X as ½ into the real equation

2x + 1/2X = 2
2½ + 1/2½ = 2
1.414 + 0.707 = 2 ( approximately)

Therefore; the answer is correct

b. simplify Log125

base ⅕

[check the picture above, to get the clearer view of the question]

(II) Prove that LOGaBASEb × LOGbBASEc × LOGcBASEa = 1

We’ll also apply the same thing to this, only that their bases will be different

LOGaBASEb × LOGbBASEc × LOGcBASEa
(LogaBASEa ÷ LogbBASEb) × (LogbBASEb ÷ LogcBASEc) ×( LogcBASEc ÷ LogaBASEa)
As we know that; Log of the same base = 1

[check the picture above, to get the clearer view of the question]

C. If Log2BASE5 = 0.431 and Log3BASE5 =0.682, evaluate Log4/9BASE5

Solution
evaluate Log4/9BASE5 = Log4BASE5 ÷ Log9BASE5
Log2²BASE5 ÷ Log3²BASE5 = 2Log2BASE5 ÷ 2Log3BASE5
USE CALCULATOR

(2 × 0.431) ÷ ( 2 × 0.682)
0.862/1.364 = 0.632( approximately)

[check the picture above, to get the clearer view of the question]