## THE DEPARTMENT OF MATHEMATICS AND STATISTICS

FIRST SEMESTER TEST 2020/2021

CLASS: ND1 civil Engineering

COURSE CODE: MTH 111

COURSE TITLE: Algebra and Element

**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

2x + 1/2X = 2

2½ + 1/2½ = 2

1.414 + 0.707 = 2 ( approximately)

##### b. simplify Log125base ⅕

**solution**

Log125base⅕ = Log125base5–¹

Let the both be in base 5

Log125base5 ÷ Log5–¹base5

Log5³base5 ÷ Log5–¹base5

3Log5base5 ÷ -1Log5base5

3÷-1 = -3

(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

1/1 × 1/1 × 1/1 = 1 × 1 × 1

=1

[check the picture above, to get the clearer view of the question]
=1

##### 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)