9th Class Mathematics Unit 2 Logarithms Numericals with Solutions

R2.2.Express the following numbers in scientific notation: (i) 0.000567 (ii) 734 (iii) 0.33 × 10^3

Result

(i) 0.000567=5.67 × 10^-4

(ii) 734=7.34 × 10²

(iii) 0.33 × 10³=3.3 × 10²

R2.3.Express the following numbers in ordinary notation: (i) 2.6 × 10^3 (ii) 8.794 × 10^-4 (iii) 6 × 10^-6

Result

(i) 2.6 × 10³=2600

(ii) 8.794 × 10^-4=0.0008794

(iii) 6 × 10^-6=0.000006

R2.4.Express each of the following in logarithmic form: (i) 3^7 = 2187 (ii) a^b = c (iii) (12)^2 = 144

Result

(i) log₃ 2187=7

(ii) log_a c=b

(iii) log₁₂144=2

R2.5.Express each of the following in exponential form: (i) log4 8 = x (ii) log9 729 = 3 (iii) log4 1024 = 5

Result

(i) 4^x=8

(ii) 9³=729

(iii) 4⁵=1024

R2.6.Find value of x in the following: (i) log9 x = 0.5 (ii) (1/9)^{3x} = 27 (iii) (1/32)^{2x} = 64

(i) log₉x=0.5 ⇒ x=9^0.5=(3²)^{12

Result

(i) x=3

(ii) left(19right)^3x=27 ⇒ (3^-2)^3x=3³ ⇒ -6x=3

(ii) x=-12

(iii) left(132right)^2x=64 ⇒ (2^-5)^2x=2⁶ ⇒ -10x=6

(iii) x=-35

R2.7.Write the following as a single logarithm: (i) 7 log x − 3 log y^2 (ii) 3 log 4 − log 32 (iii) 1/3(log5 8 + log5 27) − log5 3

Result

(i) 7log x-3log y²=logx⁷y⁶

(ii) 3log4-log32=log4³32=log6432

(ii) =log2

(iii) 13(log₅ 8+log₅ 27)-log₅ 3=13log₅(216)-log₅ 3=log₅ 6-log₅ 3

(iii) =log₅ 2

R2.8.Expand the following using laws of logarithms: (i) log(xyz^6) (ii) log3 ⁶√(m^5n^3) (iii) log√(8x^3)

Result

(i) log(xyz⁶)=log x+log y+6log z

(ii) log₃sqrt[6]{m⁵n³}=log₃(m⁵n³)^{16=16left[log₃m⁵+log₃n³right]

(ii) =16left[5log₃m+3log₃nright]

(iii) logsqrt{8x³}=log(8x³)^{12=log(2³x³)^{12=log(2x)^{32

(iii) =32[log2+log x]

R2.9.Find the values of the following with the help of logarithm table: (i) ∛68.24 (ii) 319.8 × 3.543 (iii) (36.12 × 750.9)/(113.2 × 9.98)

(i) log x=13log(68.24)=13(1.8340)=0.6113

Result

(i) x=antilog(0.6113)=4.086

(ii) log x=log(319.8)+log(3.543)=2.5049+0.5494=3.0543

(ii) x=antilog(3.0543)=1133

(iii) log x=log(36.12)+log(750.9)-log(113.2)-log(9.98)=1.5578+2.8756-2.0539-0.9991=1.3804

(iii) x=antilog(1.3804)=24.01

R2.10.In the year 2016, the population of a city was 22 millions and was growing at a rate of 2.5% per year. The function p(t) = 22(1.025)^t gives the population in millions, t years after 2016. Use the model to determine in which year the population will reach 35 millions. Round the answer to the nearest year.

Given

Model P(t)=22 × (1.025)^t

Working

Setting P(t)=35 35=22 × (1.025)^t ⇒ 1.591=(1.025)^t

Taking log log1.591=t × log1.025 ⇒ 0.2014=t × 0.0107 ⇒ t=0.20140.0107

t=18.81 ≈ 19 years

Result

Year reached Year ≈ 2016+19 ≈ 2035

Unit 2: Logarithms — Review Exercise