## Further logarithms Questions

1.  Solve for x

125-x x 52(x-2) = 25 (x+2)  (3mks)

2.  Solve for y in the equation    (3mks)

3.  Find the value of x if 49 x+1 + 7 2x = 350  (3mks)

4.  Find x in 2(4x) – 10(2x) + 8 = 0 (4mks)

5.  Solve for x in 22x – 18 x 2x = 40 (3mks)

6.  Evaluate without using mathematical tables.  (3mks)

7.  Given that log10 3 = x and log10 7 = y, Express in terms of x and y. (3mks)

8.  Find x if 32x+3 + 1 = 28 (2 mks)

9.  If ,find the value of x  (3mks)

10.  In this question, show all the steps in your calculations, giving the answer at each stage.

Use logarithms correct to 4 decimal places, to evaluate; (1934)2 x 0.00324

Log 746

11.  The table below shows monthly income tax rates

ecolebooks.com
 Monthly taxable pay in KE Rate of the tax (Kshs/ E) 1 – 342 2 343 – 684 3 685 – 1026 4 1027 – 1368 5 1369 – 1710 6 1710 and above 7

Mr. Kamau who is a civil servant earns a Monthly salary of Kshs.20000 and is provided with a house at a nominal rent of Kshs.700 per month

a) Taxable pay is the employee’s salary plus 15% less nominal rent. Calculate Mr.Kamau’s

taxable pay

b) Calculate the total tax Mr. Kamau pays

c) Mr. Kamau is entitled to a personal relief of Kshs.600 per month. What was his net tax   .

d) Mr. Kamau has the following deductions made on his pay;

Loan repayment of Kshs.2100 per month

NSSF Kshs.200 per month

WCPS calculated at 2 % of monthly salary

Calculate Mr. Kipchokes net pay

12.  A man bought a matatu at Kshs.400,000 in January 1999. It depreciated at a rate of 16%

per annum. If he valued it six months, calculate its value in January 2003

13.  The table shows corresponding values of x and y for a certain curve;

 x 1 1.2 1.4 1.6 1.8 2 2.3 y 6.5 6.2 5.2 4.3 4 2.6 2.4

Using 3 strips and mid-ordinate rule estimate the area between the curve, x-axis,

the lines x =1 and x =2.2

14.  Evaluate without using a calculator or mathematical tables.

Log 32 + log 128 – log 729

Log 32 + log 2 – log 27

15.  Find the value of x that satisfies the equation:-

log (x+5) = log 4 – log(x+2)

16.  Find the least number of terms for which the sum of the GP 100 + 200 + 400 + ………………..

exceeds 3100.

17.  A two digit number is formed from the first four prime numbers.

a) Draw the table to show the possible outcomes, if each number can be used only once.

b) Calculate the probability that a number chosen from the digit numbers is an even number

18.  Find the gradient of a line joining the centre of a circle whose equation is x2 + y2 – 6x = 3 – 4y

and a point P(6,7) outside the circle..

19.  A lady invests shs.10,000 in an account which pays 16% interest p.a. The interest is

compounded quarterly. Find the amount in the account after 1½ hrs

20.  Use logarithm tables to evaluate

13.6 cos 40o

63.5

21.  Without using logarithms or calculator evaluate:

2log105 – 3log102 + log1032

22.  Evaluate without using tables or calculators.

Log 3x + 8 – 3 log2 = log x – 4

1.

 No19342 √0.00324   2.8727 Anti log 4.8699 = 7.4114 X 10 = 74114 Log3.2865 X 2= 6.5729-3.5105 : 2= 2.7553= 5.3280.4583= 4.8699

2.  a) monthly taxable pay;

15% of monthly salary = 15/100 x 20000

= kshs.3000

Monthly pay = Kshs.(20000 + 3000 – 700)

= Kshs. 22300

In Kenya pounds = 22300/20

= KE 1115

b) Total tax payable (Gross tax)

1 – 342_________ 342×2 = Kshs.684

343 – 684 _______ 342 x3=Kshs.1026

685 – 1026 ______ 342 x4= Kshs.1368

1027 – 1368 _____ 89×5 = Kshs.445

Total tax = Kshs.3523

c) Net tax

= Gross tax – relief

= Kshs.(3523 – 600) = Kshs.2923

d) Net pay;

= Kshs.20000 – (2923 + 2100 + 200 + 2/100 x 20000)

= Kshs. (20000 – 5623) = Kshs.14377

3.  6 month depreciation rate = 8%

Number of periods = 8

400,000 (1 – 0.08)8 = 205288

4.  Mid ordinate

Area = 1.2 (6.2 + 4.3 + 2.6)

= 15.72

5.  N. log 25 x 27
= log 212

36 36

= log 22 = 4

1. 3

D; log 25 x 21 = log 26
= log 22 = log 4

33 33 33 3

N Log 4

D 3

Log 4

3

= 6 log 4

3

3 log 4

3

6/3 = 2

6.   Log (x+5) = log(4)

(x+2)

x + 5 = 4

x + 2

(x+5) (x+2) = 4

x2 + 2x + 5X + 10 = 4

x2 + 7x + 6 = 0

x2 + 6x + x + 6 =0

x(x+6) + 1(x+6) = 0

(x +1) (x +6) = 0

x = -1 x = -6

7.  a = 100

r = 200 = 2

100

a (rn -1) ≻Sn

r – 1

100 (2n – 1) ≻ 3,100

2 – 1

2n – 1  31

2n
 32

2n
 25

n  5

n = 6

8. a)   2   3 5  7

2 32  52  72

3  23 53   73

5  25  35 75

7  27   37  57

b) P(E) = 4

16

= ¼

9.   x2 + y2 – 6x = 3 – 4y

x2 – 6x + (-6/2)2 + y2 + 4y + (4/2)2 = 3 + (-6/2)2 + (4/2)2

(x – 3)2 (y + 2)2 = 3 + 9 = 4

(x – 3)2 (y + 2)2 = 16

C (3, -2)

Gradient ∆y = 7 – -2 = 3

∆x 6 – 3

10.  A = P(1 + r )n

100

= 10000 (1 + 4 )6

100

= 10000(1.04)6

= 12653.19 (12,653)

11. No. Std. Form Log

13.6 1.36 x 101 1.1335

+

Cos 400 1.8842

1.0177

__

63.5 6.35 x 101 1.8028

1.2149 ÷3

= 3 + 2.2149

33

0.5474 5.474 x 10-1 1.7383

0.5474

12.  Log10 52 – log10 23 + log 25

Log10
25 x 32

8

Log10 100 = log1010

= 2 log 1010

But log1010 = 1

∴ = 2

13.  Log 3x + 8 = log (x -4)

23

3x +8 = x – 4

8

3x + 8 = 8(x – 4)

3x + 8 = 8x – 32

– 5x = -40

x = 8

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## 1 Comment

• ### Becca, August 13, 2022 @ 1:54 pmReply

I can’t find the answers to some of the questions .Can you please make it a bit organised because it’s so tiring searching for the answers, anyways it’s a really great app

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