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Class 9 Mathematics Notes — Chapter 2: Logarithms

Punjab Board · Class 9 · Chapter 02

Exercise 2.1

Q1. Express the following numbers in scientific notation.

(i) \(2,000,000\)

\[ 2,000,000=2\times10^6 \]

Move the decimal point 6 places to the left.

(ii) \(48,900\)

\[ 48,900=4.89\times10^4 \]

Move the decimal point 4 places to the left.

(iii) \(0.0042\)

\[ 0.0042=4.2\times10^{-3} \]

Move the decimal point 3 places to the right.

(iv) \(0.000009\)

\[ 0.000009=9\times10^{-6} \]

Move the decimal point 6 places to the right.

(v) \(73\times10^3\)

\[ 73\times10^3=7.3\times10^1\times10^3 \]

\[ \boxed{7.3\times10^4} \]

(vi) \(0.65\times10^2\)

\[ 0.65\times10^2=6.5\times10^{-1}\times10^2 \]

\[ \boxed{6.5\times10^1} \]

Q2. Express the following numbers in ordinary notation.

(i) \(8.04\times10^2\)

\[ \boxed{804} \]

(ii) \(3\times10^5\)

\[ \boxed{300,000} \]

(iii) \(1.5\times10^{-2}\)

\[ \boxed{0.015} \]

(iv) \(1.77\times10^7\)

\[ \boxed{17,700,000} \]

(v) \(5.5\times10^{-6}\)

\[ \boxed{0.0000055} \]

(vi) \(4\times10^{-5}\)

\[ \boxed{0.00004} \]

Q3. The speed of light is approximately \(3\times10^8\) metres per second. Express it in standard form.

Solution:

The number is already in standard scientific notation.

\[ \boxed{3\times10^8\text{ m/s}} \]

Q4. The circumference of the Earth at the equator is about \(40,075,000\) metres. Express this number in scientific notation.

Solution:

Move the decimal point 7 places to the left.

\[ \boxed{4.0075\times10^7\text{ m}} \]

Q5. The diameter of Mars is \(6.7779\times10^3\) km. Express this number in standard form.

Solution:

The number is already in standard form.

\[ \boxed{6.7779\times10^3\text{ km}} \]

Q6. The diameter of Earth is about \(1.2756\times10^4\) km. Express this number in standard form.

Solution:

The number is already in standard form.

\[ \boxed{1.2756\times10^4\text{ km}} \]

Exercise 2.2

Q1. Express each of the following in logarithmic form.

(i) \(10^3=1000\)

\[ \boxed{\log_{10}1000=3} \]

(ii) \(2^8=256\)

\[ \boxed{\log_2 256=8} \]

(iii) \(3^{-3}=\dfrac{1}{27}\)

\[ \boxed{\log_3\frac{1}{27}=-3} \]

(iv) \(20^2=400\)

\[ \boxed{\log_{20}400=2} \]

(v) \(16^{-\frac14}=\dfrac12\)

\[ \boxed{\log_{16}\frac12=-\frac14} \]

(vi) \(11^2=121\)

\[ \boxed{\log_{11}121=2} \]

(vii) \(p=q^r\)

\[ \boxed{\log_q p=r} \]

(viii) \(32^{-\frac15}=\dfrac12\)

\[ \boxed{\log_{32}\frac12=-\frac15} \]

Q2. Express each of the following in exponential form.

(i) \(\log_5 125=3\)

\[ \boxed{5^3=125} \]

(ii) \(\log_2 16=4\)

\[ \boxed{2^4=16} \]

(iii) \(\log_{23}1=0\)

\[ \boxed{23^0=1} \]

(iv) \(\log_x5=1\)

\[ \boxed{x^1=5} \]

(v) \(\log_2\frac18=-3\)

\[ \boxed{2^{-3}=\frac18} \]

(vi) \(\dfrac12=\log_2 3\)

The supplied question appears to have its logarithm written in reverse order. Following the given conversion:

\[ \boxed{2^{1/2}=\sqrt2} \]

This does not equal \(\sqrt3\); please check the original question.

(vii) \(5=\log_{10}100,000\)

\[ \boxed{10^5=100,000} \]

(viii) \(\log_4\frac1{16}=-2\)

\[ \boxed{4^{-2}=\frac1{16}} \]

Q3. Find the value of \(x\) in each of the following.

(i) \(\log_x64=3\)

\[ x^3=64=4^3 \]

\[ \boxed{x=4} \]

(ii) \(\log_5 1=x\)

\[ 5^x=1=5^0 \]

\[ \boxed{x=0} \]

(iii) \(\log_x8=1\)

\[ x^1=8 \]

\[ \boxed{x=8} \]

(iv) \(\log_{10}x=-3\)

\[ x=10^{-3} \]

\[ \boxed{x=0.001} \]

(v) \(\log_4x=\dfrac32\)

\[ x=4^{3/2}=(2^2)^{3/2}=2^3 \]

\[ \boxed{x=8} \]

(vi) \(\log_2 1024=x\)

\[ 2^x=1024=2^{10} \]

\[ \boxed{x=10} \]

Exercise 2.3

Q1. Find the characteristic of the following numbers.

(i) \(5287\)

There are 4 digits before the decimal point.

\[ \text{Characteristic}=4-1 \]

\[ \boxed{3} \]

(ii) \(59.28\)

There are 2 digits before the decimal point.

\[ \text{Characteristic}=2-1 \]

\[ \boxed{1} \]

(iii) \(0.0567\)

The first non-zero digit appears at the second decimal place.

\[ \boxed{\overline{2}} \]

(iv) \(234.7\)

There are 3 digits before the decimal point.

\[ \text{Characteristic}=3-1 \]

\[ \boxed{2} \]

(v) \(0.000049\)

The first non-zero digit appears at the fifth decimal place.

\[ \boxed{\overline{5}} \]

(vi) \(145,000\)

There are 6 digits before the decimal point.

\[ \text{Characteristic}=6-1 \]

\[ \boxed{5} \]

Q2. Find the logarithm of the following numbers.

(i) \(43\)

Characteristic \(=1\), mantissa \(=0.6335\).

\[ \boxed{\log 43=1.6335} \]

(ii) \(579\)

Characteristic \(=2\), mantissa \(=0.7627\).

\[ \boxed{\log 579=2.7627} \]

(iii) \(1.982\)

Characteristic \(=0\), mantissa \(=0.2971\).

\[ \boxed{\log 1.982=0.2971} \]

(iv) \(0.0876\)

Characteristic \(=\overline{2}\), mantissa \(=0.9425\).

\[ \boxed{\log 0.0876=\overline{2}.9425} \]

(v) \(0.047\)

Characteristic \(=\overline{2}\), mantissa \(=0.6721\).

\[ \boxed{\log 0.047=\overline{2}.6721} \]

(vi) \(0.000354\)

Characteristic \(=\overline{4}\), mantissa \(=0.5490\).

\[ \boxed{\log 0.000354=\overline{4}.5490} \]

Q3. If \(\log 3.177=0.5019\), then find the following.

(i) \(\log 3177\)

Characteristic \(=3\).

\[ \boxed{\log 3177=3.5019} \]

(ii) \(\log 31.77\)

Characteristic \(=1\).

\[ \boxed{\log 31.77=1.5019} \]

(iii) \(\log 0.03177\)

Characteristic \(=\overline{2}\).

\[ \boxed{\log 0.03177=\overline{2}.5019} \]

Q4. Find the value of \(x\).

(i) \(\log x=0.0065\)

\[ x=\operatorname{antilog}(0.0065) \]

\[ \boxed{x\approx1.015} \]

(ii) \(\log x=1.192\)

\[ x=\operatorname{antilog}(1.192) \]

\[ \boxed{x\approx15.56} \]

(iii) \(\log x=-3.434\)

Expressing the logarithm with a negative characteristic:

\[ -3.434=\overline{4}.566 \]

\[ x=\operatorname{antilog}(\overline{4}.566) \]

\[ \boxed{x\approx0.000368} \]

(iv) \(\log x=-1.5726\)

\[ -1.5726=\overline{2}.4274 \]

\[ x=\operatorname{antilog}(\overline{2}.4274) \]

\[ \boxed{x\approx0.02675} \]

(v) \(\log x=4.3561\)

\[ x=\operatorname{antilog}(4.3561) \]

\[ \boxed{x\approx22,700} \]

(vi) \(\log x=-2.0184\)

\[ -2.0184=\overline{3}.9816 \]

\[ x=\operatorname{antilog}(\overline{3}.9816) \]

\[ \boxed{x\approx0.009585} \]

Exercise 2.4

Q1. Without using a calculator, evaluate the following.

(i) \(\log_2 18-\log_2 9\)

Using the quotient law:

\[ \log_2\frac{18}{9}=\log_2 2 \]

\[ \boxed{1} \]

(ii) \(\log_2 64+\log_2 2\)

Using the product law:

\[ \log_2(64\times2)=\log_2 128 \]

\[ =\log_2 2^7 \]

\[ \boxed{7} \]

(iii) \(\dfrac13\log_3 8-\log_3 18\)

Using the power law:

\[ \log_3 8^{1/3}-\log_3 18 \]

\[ =\log_3 2-\log_3 18 \]

\[ =\log_3\frac2{18}=\log_3\frac19 \]

\[ \boxed{-2} \]

(iv) \(2\log 2+\log 5\)

\[ =\log 2^2+\log 5 \]

\[ =\log(4\times5)=\log20 \]

The supplied solution gives \(\log100=2\), but that would require \(\log 4+\log25\), not \(\log4+\log5\).

\[ \boxed{\log20} \]

(v) \(\dfrac13\log_5 64+2\log_5 25\)

\[ =\log_5 4+\log_5 625 \]

\[ =\log_5(4\times625) \]

\[ \boxed{\log_5 2500} \]

(vi) \(\log_3 12+\log_3 0.25\)

\[ =\log_3(12\times0.25) \]

\[ =\log_3 3 \]

\[ \boxed{1} \]

Q2. Write the following as a single logarithm.

(i) \(\dfrac12\log_2 5+2\log_3\)

The second logarithm’s base and argument are unclear in the supplied text. The provided working gives:

\[ \log 5+\log9 \]

\[ \boxed{\log45} \]

Please check the original expression to confirm the bases.

(ii) \(\log9-\log\left(\dfrac13\right)\)

\[ =\log\left(\frac9{1/3}\right) \]

\[ =\log(9\times3) \]

\[ \boxed{\log27} \]

(iii) \(\log_a b^3-\log_a 5^3\)

\[ =\log_a\left(\frac{b^3}{5^3}\right) \]

\[ \boxed{\log_a\frac{b^3}{125}} \]

(iv) \(2\log_3x+\log_3y\)

\[ =\log_3x^2+\log_3y \]

\[ \boxed{\log_3(x^2y)} \]

(v) \(4\log_5x-\log_5y+\log_5z\)

\[ =\log_5x^4-\log_5y+\log_5z \]

\[ \boxed{\log_5\frac{x^4z}{y}} \]

(vi) \(2\ln a+3\ln b-4\ln c\)

\[ =\ln a^2+\ln b^3-\ln c^4 \]

\[ \boxed{\ln\frac{a^2b^3}{c^4}} \]

Q3. Expand the following using laws of logarithms.

(i) \(\log\left(\dfrac{11}{5}\right)\)

\[ \boxed{\log11-\log5} \]

(ii) \(\log_5\sqrt{8a^5}\)

\[ =\frac12\log_5(8a^5) \]

\[ =\frac12(\log_5 8+5\log_5a) \]

\[ \boxed{\frac12\log_5 8+\frac52\log_5a} \]

(iii) \(\ln\left(\dfrac{a^2b}{c}\right)\)

\[ \boxed{2\ln a+\ln b-\ln c} \]

(iv) \(\log\left(\dfrac{xy}{z}\right)^3\)

\[ =3\log\left(\frac{xy}{z}\right) \]

\[ \boxed{3\log x+3\log y-3\log z} \]

(v) \(\ln\sqrt[3]{16x^3}\)

\[ =\frac13\ln(16x^3) \]

\[ =\frac13\ln16+\ln x \]

\[ \boxed{\frac13\ln16+\ln x} \]

(vi) \(\log_2\left(\dfrac{1-a}{b}\right)^5\)

\[ =5\log_2\left(\frac{1-a}{b}\right) \]

\[ \boxed{5\log_2(1-a)-5\log_2b} \]

Q4. Find the value of \(x\) in the following equations.

(i) \(\log2+\log x=1\)

\[ \log(2x)=1=\log10 \]

\[ 2x=10 \]

\[ \boxed{x=5} \]

(ii) \(\log_2x+\log_2 8=5\)

\[ \log_2(8x)=5 \]

\[ 8x=2^5=32 \]

\[ \boxed{x=4} \]

(iii) \(81^x=243^{x+2}\)

\[ (3^4)^x=(3^5)^{x+2} \]

\[ 4x=5(x+2) \]

\[ 4x=5x+10 \]

\[ \boxed{x=-10} \]

(iv) \(\left(\dfrac1{27}\right)^{x-6}=27\)

\[ (3^{-3})^{x-6}=3^3 \]

\[ -3(x-6)=3 \]

\[ -3x+18=3 \]

\[ \boxed{x=5} \]

(v) \(\log(5x-10)=2\)

\[ 5x-10=10^2 \]

\[ 5x-10=100 \]

\[ 5x=110 \]

\[ \boxed{x=22} \]

(vi) \(\log_2(x+1)-\log_2(x-4)=2\)

\[ \log_2\left(\frac{x+1}{x-4}\right)=2 \]

\[ \frac{x+1}{x-4}=2^2=4 \]

\[ x+1=4(x-4) \]

\[ x+1=4x-16 \]

\[ 17=3x \]

\[ \boxed{x=\frac{17}{3}} \]

Q5. Find the values using logarithm tables.

(i) \(\dfrac{3.68\times4.21}{5.234}\)

\[ \boxed{2.961} \]

(ii) \(4.67\times2.11\times2.397\)

\[ \boxed{23.61} \]

(iii) \(\dfrac{(20.46)^2\times(2.412)^2}{754.3}\)

\[ \boxed{1.338} \]

(iv) \(\sqrt[3]{9.364\times21.64}\div3.21\)

\[ \boxed{0.867} \]

Note: These numerical answers are retained from the supplied notes; the original expressions should be checked against the textbook if the results differ.

Q6. The formula to measure the magnitude of earthquakes is \(M=\log_{10}\left(\dfrac{A}{A_0}\right)\). If \(A=10,000A_0\), find \(M\).

Solution:

\[ M=\log_{10}\left(\frac{10,000A_0}{A_0}\right) \]

\[ =\log_{10}(10,000) \]

\[ =\log_{10}(10^4) \]

\[ \boxed{M=4} \]

Q7. Abdullah invested Rs. 100,000 at 5% per annum so that \(y=100,000(1.05)^t\). Find after how many years the investment will double.

Solution:

For the investment to double:

\[ 200,000=100,000(1.05)^t \]

Dividing both sides by \(100,000\):

\[ (1.05)^t=2 \]

Taking logarithms:

\[ t\log1.05=\log2 \]

\[ t=\frac{\log2}{\log1.05} \]

\[ \boxed{t\approx14.2\text{ years}} \]

Q8. Huria is hiking a mountain where temperature decreases by 3% per 100 m. Given \(T_s=20^\circ C\) and \(h=500\) m, calculate the temperature.

Solution:

The temperature decreases by 3% for every 100 m.

For 500 m, there are five 100 m intervals.

\[ T=20(0.97)^5 \]

\[ T\approx17.17^\circ C \]

\[ \boxed{T\approx17.17^\circ C} \]

Review Exercise 2

Q1. Multiple Choice Questions

(i) The standard form of \(5.2\times10^6\) is:

Correct Answer: (c) 5,200,000

(ii) Scientific notation of \(0.00034\) is:

Correct Answer: (b) \(3.4\times10^{-4}\)

(iii) The base of common logarithm is:

Correct Answer: (b) 10

(iv) \(\log_2 2^3=\)

Correct Answer: (d) 3

(v) \(\log100=\)

Correct Answer: (a) 2

(vi) If \(\log2=0.3010\), then \(\log200\) is:

\[ \log200=\log(2\times100)=\log2+\log100 \]

\[ =0.3010+2=2.3010 \]

Correct Answer: (c) 2.3010

(vii) \(\log(0)=\)

Correct Answer: (d) Undefined

(viii) \(\log10,000=\)

Correct Answer: (c) 4

(ix) \(\log5+\log3=\)

\[ =\log(5\times3) \]

Correct Answer: (d) \(\log15\)

(x) \(3^4=81\) in logarithmic form is:

Correct Answer: (c) \(\log_3 81=4\)

Q2. Express the following numbers in scientific notation.

(i) \(0.000567\)

\[ \boxed{5.67\times10^{-4}} \]

(ii) \(734\)

\[ \boxed{7.34\times10^2} \]

(iii) \(0.33\times10^3\)

\[ =3.3\times10^{-1}\times10^3 \]

\[ \boxed{3.3\times10^2} \]

Q3. Express the following numbers in ordinary notation.

(i) \(2.6\times10^3\)

\[ \boxed{2,600} \]

(ii) \(8.794\times10^{-4}\)

\[ \boxed{0.0008794} \]

(iii) \(6\times10^6\)

\[ \boxed{6,000,000} \]

Q4. Express each of the following in logarithmic form.

(i) \(3^7=2187\)

\[ \boxed{\log_3 2187=7} \]

(ii) \(a^b=c\)

\[ \boxed{\log_a c=b} \]

(iii) \(12^2=144\)

\[ \boxed{\log_{12}144=2} \]

Q5. Express each of the following in exponential form.

(i) \(\log_4 8=x\)

\[ \boxed{4^x=8} \]

(ii) \(\log_3 729=3\)

The supplied equation is inconsistent because \(3^3=27\), not \(729\). If the base is 3 and the argument is 729, then:

\[ \boxed{3^6=729} \]

Please verify the original question.

(iii) \(\log_4 1024=5\)

\[ \boxed{4^5=1024} \]

Q6. Find the value of \(x\) in the following.

(i) \(\log_9x=0.5\)

\[ x=9^{0.5}=\sqrt9 \]

\[ \boxed{x=3} \]

(ii) \(\left(\dfrac19\right)^{3x}=27\)

\[ (3^{-2})^{3x}=3^3 \]

\[ -6x=3 \]

\[ \boxed{x=-\frac12} \]

(iii) \(\left(\dfrac1{32}\right)^{2x}=64\)

\[ (2^{-5})^{2x}=2^6 \]

\[ -10x=6 \]

\[ \boxed{x=-\frac35} \]

Q7. Write the following as a single logarithm.

(i) \(7\log x-3\log y^2\)

\[ =7\log x-6\log y \]

\[ \boxed{\log\left(\frac{x^7}{y^6}\right)} \]

(ii) \(3\log4-\log32\)

\[ =\log4^3-\log32 \]

\[ =\log\left(\frac{64}{32}\right) \]

\[ \boxed{\log2} \]

(iii) \(\dfrac13(\log_5 8+\log_5 27)-\log_5 3\)

\[ =\log_5\sqrt[3]{8\times27}-\log_5 3 \]

\[ =\log_5 6-\log_5 3 \]

\[ \boxed{\log_5\frac23} \]

Q8. Expand the following using laws of logarithms.

(i) \(\log(xyz^6)\)

\[ \boxed{\log x+\log y+6\log z} \]

(ii) \(\log_3\sqrt[5]{m^4n^3}\)

\[ =\frac15\log_3(m^4n^3) \]

\[ \boxed{\frac45\log_3m+\frac35\log_3n} \]

(iii) \(\log\sqrt{8x^3}\)

\[ =\frac12\log(8x^3) \]

\[ =\frac12\log8+\frac32\log x \]

Since \(\log8=3\log2\):

\[ \boxed{\frac32\log2+\frac32\log x} \]

Q9. Find the values using logarithm tables.

(i) \(\sqrt[3]{68.24}\)

\[ \boxed{4.087} \]

(ii) \(319.8\times3.543\)

\[ \boxed{1133} \]

(iii) \(\dfrac{36.12\times750.9}{113.2\times9.98}\)

\[ \boxed{24.01} \]

The answers above are retained from the supplied notes; verify the original expressions if your textbook gives different values.

Q10. In the year 2016, population \(p(t)=22(1.025)^t\) millions. Determine in which year the population will reach 35 million.

Solution:

Given:

\[ 35=22(1.025)^t \]

Dividing both sides by 22:

\[ (1.025)^t=\frac{35}{22} \]

\[ (1.025)^t\approx1.5909 \]

Taking logarithms:

\[ t=\frac{\log(1.5909)}{\log(1.025)} \]

\[ t\approx18.8\text{ years} \]

Therefore:

\[ 2016+18.8\approx2035 \]

\[ \boxed{\text{Year }2035} \]

Notes by Sir Dilawar

For more notes, visit QuranEducationPoint.com