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

## THE DEPARTMENT OF MATHEMATICS AND STATISTICSFIRST SEMESTER TEST 2020/2021CLASS: ND1 civil EngineeringCOURSE CODE: MTH 111COURSE 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)
##### 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
[check the picture above, to get the clearer view of the question]
##### 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]
##### 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]

### Oluwamuyide Peter

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