9th Class Mathematics Unit 2 Logarithms Numericals with Solutions

2.4.1.Without using calculator, evaluate the following: (i) log2 18 − log2 9 (ii) log2 64 + log2 2 (iii) 1/3 log3 8 − log3 18 (iv) 2log 2 + log 25 (v) 1/3 log4 64 + 2log5 25 (vi) log3 12 + log3 0.25

(i) log₂ 18-log₂ 9=log₂(2 × 9)-log₂ 9=log₂ 2+log₂ 9-log₂ 9=log₂ 2

Result

(i) =1

(ii) log₂ 64+log₂ 2=log₂(2⁶)+log₂ 2=6log₂ 2+log₂ 2=7log₂ 2

(ii) =7

(iii) 13log₃ 8-log₃ 18=13log₃(2³)-log₃(2 × 3²)=log₃ 2-log₃ 2-2log₃ 3

(iii) =-2

(iv) 2log 2+log 25=2log 2+2log 5=2(log 2+log 5)=2log 10

(iv) =2

(v) 13log₄ 64+2log₅ 25=13log₄(4³)+2log₅(5²)=log₄ 4+4log₅ 5=1+4

(v) =5

(vi) log₃ 12+log₃ 0.25=log₃ 12+log₃14=log₃124=log₃ 3

(vi) =1

2.4.2.Write the following as a single logarithm: (i) 1/2 log 25 + 2 log 3 (ii) log 9 − log 1/3 (iii) log5 b^2 . log_a 5^3 (iv) 2log3 x + log3 y (v) 4log5 x − log5 y + log5 z (vi) 2 ln a + 3 ln b − 4 ln c

(i) 12log25+2log3=log5+log9

Result

(i) =log45

(ii) log9-log13=logleft(9 × 3right)

(ii) =log27

(iii) log₅b² · log_a5³=2log₅b × 3log_a5=6log_a blog_a5 × log_a5

(iii) =6log_a b

(v) 4log₅x-log₅y+log₅z = log₅x⁴zy

The book's printed solution has no working for this part at all — it is skipped between parts (iii) and (vi). The correct combined form is log5(x^4 z / y).

(iv), mislabelled "vi." in source 2log₃x+log₃y=log₃(x²)+log₃y

(iv) =log₃x²y

(vi) 2ln a+3ln b-4ln c=ln a²+ln b³-ln c⁴

(vi) =lna²b³c⁴

2.4.3.Expand the following using laws of logarithms: (i) log(11/5) (ii) log5 √(8a^6) (iii) ln(a^2 b/c) (iv) log((xy/z)^{1/9}) (v) ln ∛(16x^3) (vi) log2(((1−a)/b)^5)

Result

(i) logleft(115right)=log11-log5

(ii) log₅sqrt{8a⁶}=log₅(2³ × a⁶)^{12=log₅left(2^{32 × a³right)

(ii) =32log₅ 2+3log₅ a

(iii) lnleft(a²bcright)=ln a²+ln b-ln c=2ln a+ln b-ln c

(iv) logleft(xyzright)^{19=19left[log x+log y-log zright]

Book's printed working switches notation to ln instead of log for this part; the expansion law applied is otherwise correct.

(v) lnsqrt[3]{16x³}=ln(2⁴ × x³)^{13=lnleft(2^{43 × xright)

(v) =43ln2+ln x

(vi) log₂left(1-abright)⁵=5log₂left(1-abright)=5left[log₂(1-a)-log₂ bright]

2.4.4.Find the value of x in the following equations: (i) log 2 + log x = 1 (ii) log2 x + log2 8 = 5 (iii) (81)^x = (243)^{x+2} (iv) (1/27)^{x−6} = 27 (v) log(5x−10) = 2 (vi) log2(x+1) − log2(x−4) = 2

(i) log2+log x=1 ⇒ log2x=log10 ⇒ 2x=10

Result

(i) x=5

(ii) log₂x+log₂8=5 ⇒ log₂ 8x=log₂ 2⁵ ⇒ 8x=32

(ii) x=4

(iii) (81)^x=(243)^x+2 ⇒ (3⁴)^x=(3⁵)^x+2 ⇒ 4x=5x+10

(iii) x=-10

(iv) left(127right)^x-6=27 ⇒ (3^-3)^x-6=3³ ⇒ -3x+18=3

(iv) x=5

(v) log(5x-10)=2 ⇒ 5x-10=10² ⇒ 5x=110

(v) x=22

(vi) log₂(x+1)-log₂(x-4)=2 ⇒ x+1x-4=2² ⇒ x+1=4x-16 ⇒ 3x=17

(vi) x=173=523

2.4.5.Find the values of the following with the help of logarithm table: (i) (3.68 × 4.21)/5.234 (ii) 4.67 × 2.11 × 2.397 (iii) (20.46)^2 × 2.4122 / 754.3 (iv) ∛9.364 × 21.64 / 3.21

(i) log x=log(3.68)+log(4.21)-log(5.234)=0.5658+0.6243-0.7188=0.4713

Result

(i) x=antilog(0.4713)=2.960

(ii) log x=log(4.67)+log(2.11)+log(2.397)=0.6693+0.3243+0.3797=1.3733

(ii) x=antilog(1.3733)=23.62

(iii) log x=2log(20.46)+log(2.4122)-log(754.3)=2(1.3109)+0.3824-2.8776=0.1266

(iii) x=antilog(0.1266)=1.339

(iv) log x=13log(9.364)+log(21.64)-log(3.21)=13(0.9715)+1.3353-0.5065=1.1526

(iv) x=antilog(1.1526)=14.21

2.4.6.The formula to measure the magnitude of earthquakes is given by M = log10(A/A0). If amplitude (A) is 10,000 and reference amplitude (A0) is 10. What is the magnitude of the earthquake?

Formula

Magnitude formula M=log₁₀left(AA₀right)

Working

Substituting M=log₁₀left(1000010right)=log₁₀(1000)=log₁₀(10³)=3log₁₀10

Result

Magnitude M=3 (rector scale)

2.4.7.Abdullah invested Rs. 100,000 in a saving scheme and gains interest at the rate of 5% per annum so that the total value of this investment after t years is Rs y. This is modelled by an equation y = 100,000 (1.05)^t, t ≥ 0. Find after how many years the investment will be double.

Given

Model y=100000 × (1.05)^t

Working

Doubling condition 200000=100000 × (1.05)^t ⇒ 2=(1.05)^t

Taking log log2=t × log(1.05) ⇒ t=log2log1.05=0.30100.0212

Result

Time to double t ≈ 14.21 years

2.4.8.Huria is hiking up a mountain where the temperature (T) decreases by 3% (or a factor of 0.97) for every 100 metres gained in altitude. The initial temperature (Ti) at sea level is 20°C. Using the formula T = Ti × 0.97^(h/100), calculate the temperature at an altitude (h) of 500 metres.

Given

Formula T=T_i × (0.97)^{h100

Working

Substituting h=500, Ti=20 T=20 × (0.97)^{500100=20 × (0.97)⁵

First method T=20 × 0.859

Second method (log) log T=log20+5log(0.97)=1.3011+5(-0.0133)=1.2346 ⇒ T=antilog(1.2346)

Result

Temperature T ≈ 17.17°C

Unit 2: Logarithms — Exercise 2 4