9th Class Math Chapter 2 Notes

Q1(i) — Scientific notation: $2000000$

Express $2000000$ in scientific notation.

1. Move the decimal point so only one non-zero digit remains before it: $2000000=2\times 10^6$.

Answer: $2\times 10^6$

Q1(ii) — Scientific notation: $48900$

Express $48900$ in scientific notation.

1. $48900=4.89\times 10^4$.

Answer: $4.89\times 10^4$

Q1(iii) — Scientific notation: $0.0042$

Express $0.0042$ in scientific notation.

1. $0.0042=4.2\times 10^{-3}$.

Answer: $4.2\times 10^{-3}$

Q1(iv) — Scientific notation: $0.0000009$

Express $0.0000009$ in scientific notation.

1. $0.0000009=9\times 10^{-7}$.

Answer: $9\times 10^{-7}$

Q1(v) — Scientific notation: $73\times 10^3$

Express $73\times 10^3$ in scientific notation.

1. The mantissa must be between $1$ and $10$: $73=7.3\times 10^1$.
2. So $73\times 10^3=7.3\times 10^{1+3}=7.3\times 10^4$.

Answer: $7.3\times 10^4$

Q1(vi) — Scientific notation: $0.65\times 10^2$

Express $0.65\times 10^2$ in scientific notation.

1. $0.65=6.5\times 10^{-1}$.
2. So $0.65\times 10^2=6.5\times 10^{-1+2}=6.5\times 10^1$.

Answer: $6.5\times 10^1$

Q1(vii) — Scientific notation: $0.000567$

Express $0.000567$ in scientific notation.

1. $0.000567=5.67\times 10^{-4}$.

Answer: $5.67\times 10^{-4}$

Q1(viii) — Scientific notation: $734$

Express $734$ in scientific notation.

1. $734=7.34\times 10^2$.

Answer: $7.34\times 10^2$

Q1(ix) — Scientific notation: $0.33\times 10^3$

Express $0.33\times 10^3$ in scientific notation.

1. $0.33=3.3\times 10^{-1}$.
2. So $0.33\times 10^3=3.3\times 10^{-1+3}=3.3\times 10^2$.

Answer: $3.3\times 10^2$

Q2(i) — Ordinary notation: $8.04\times 10^2$

Express $8.04\times 10^2$ in ordinary notation.

1. Move the decimal point $2$ places right: $8.04\to 804$.

Answer: $804$

Q2(ii) — Ordinary notation: $3\times 10^5$

Express $3\times 10^5$ in ordinary notation.

1. Move the decimal point $5$ places right: $3\to 300000$.

Answer: $300000$

Q2(iii) — Ordinary notation: $1.5\times 10^{-2}$

Express $1.5\times 10^{-2}$ in ordinary notation.

1. Move the decimal point $2$ places left: $1.5\to 0.015$.

Answer: $0.015$

Q2(iv) — Ordinary notation: $1.77\times 10^7$

Express $1.77\times 10^7$ in ordinary notation.

1. Move the decimal point $7$ places right: $1.77\to 17700000$.

Answer: $17700000$

Q2(v) — Ordinary notation: $5.5\times 10^{-6}$

Express $5.5\times 10^{-6}$ in ordinary notation.

1. Move the decimal point $6$ places left: $5.5\to 0.0000055$.

Answer: $0.0000055$

Q2(vi) — Ordinary notation: $4\times 10^{-5}$

Express $4\times 10^{-5}$ in ordinary notation.

1. Move the decimal point $5$ places left: $4\to 0.00004$.

Answer: $0.00004$

Q2(vii) — Ordinary notation: $2.6\times 10^3$

Express $2.6\times 10^3$ in ordinary notation.

1. Move the decimal point $3$ places right: $2.6\to 2600$.

Answer: $2600$

Q2(viii) — Ordinary notation: $8.794\times 10^{-4}$

Express $8.794\times 10^{-4}$ in ordinary notation.

1. Move the decimal point $4$ places left: $8.794\to 0.0008794$.

Answer: $0.0008794$

Q2(ix) — Ordinary notation: $6\times 10^{-6}$

Express $6\times 10^{-6}$ in ordinary notation.

1. Move the decimal point $6$ places left: $6\to 0.000006$.

Answer: $0.000006$

WP1 — Speed of light in standard/ordinary form

The speed of light is approximately $3\times 10^8$ m/s. Express it in standard (scientific) form and in ordinary form.

1. It is already correctly in standard/scientific form since $1\le 3<10$: $3\times 10^8$.
2. In ordinary form, move the decimal $8$ places right.

Answer: Standard form: $3\times 10^8$ m/s; ordinary form: $300,000,000$ m/s

WP2 — Earth's circumference in scientific notation

Earth's equatorial circumference is about $40,075,000$ m. Express it in scientific notation.

1. Keep one non-zero digit before the decimal: $4.0075\times 10^7$.

Answer: $4.0075\times 10^7$ m

WP3 — Diameter of Mars in ordinary form

Mars's diameter is $6.779\times 10^3$ km. Express it in ordinary form.

1. Move the decimal $3$ places right: $6.779\to 6779$.

Answer: $6779$ km

WP4 — Diameter of Earth in ordinary form

Earth's diameter is about $1.2756\times 10^4$ km. Express it in ordinary form.

1. Move the decimal $4$ places right: $1.2756\to 12756$.

Answer: $12756$ km

Q3(i) — Logarithmic form: $10^3=1000$

Write $10^3=1000$ in logarithmic form.

1. Base $\to$ base of the log; exponent $\to$ value of the log.

Answer: ${\log }_{10}1000=3$

Q3(ii) — Logarithmic form: $2^8=256$

Write $2^8=256$ in logarithmic form.

1. Base $2$, exponent $8$.

Answer: ${\log }_{2}256=8$

Q3(iii) — Logarithmic form: $3^{-3}=\frac{1}{27}$

Write $3^{-3}=\frac{1}{27}$ in logarithmic form.

1. Base $3$, exponent $-3$.

Answer: ${\log }_3\frac{1}{27}=-3$

Q3(iv) — Logarithmic form: $20^2=400$

Write $20^2=400$ in logarithmic form.

1. Base $20$, exponent $2$.

Answer: ${\log }_{20}400=2$

Q3(v) — Logarithmic form: $16^{-\frac{1}{4}}=\frac{1}{2}$

Write $16^{-\frac{1}{4}}=\frac{1}{2}$ in logarithmic form.

1. Base $16$, exponent $-\frac{1}{4}$.

Answer: ${\log }_{16}\frac{1}{2}=-\frac{1}{4}$

Q3(vi) — Logarithmic form: $11^2=121$

Write $11^2=121$ in logarithmic form.

1. Base $11$, exponent $2$.

Answer: ${\log }_{11}121=2$

Q3(vii) — Logarithmic form: $p=q^r$

Write $p=q^r$ in logarithmic form.

1. Base $q$, value of exponent $r$, result $p$.

Answer: ${\log }_{q}p=r$

Q3(viii) — Logarithmic form: $(32{)}^{-\frac{1}{5}}=\frac{1}{2}$

Write $(32{)}^{-\frac{1}{5}}=\frac{1}{2}$ in logarithmic form.

1. Base $32$, exponent $-\frac{1}{5}$.

Answer: ${\log }_{32}\frac{1}{2}=-\frac{1}{5}$

Q3(ix) — Logarithmic form: $3^7=2187$

Write $3^7=2187$ in logarithmic form.

1. Base $3$, exponent $7$.

Answer: ${\log }_{3}2187=7$

Q3(x) — Logarithmic form: $a^b=c$

Write $a^b=c$ in logarithmic form.

1. Base $a$, exponent $b$, result $c$.

Answer: ${\log }_{a}c=b$

Q3(xi) — Logarithmic form: $(12{)}^2=144$

Write $(12{)}^2=144$ in logarithmic form.

1. Base $12$, exponent $2$.

Answer: ${\log }_{12}144=2$

Q4(i) — Exponential form: ${\log }_{5}125=3$

Write ${\log }_{5}125=3$ in exponential form.

1. Base raised to the log value equals the number: $5^3=125$.

Answer: $5^3=125$

Q4(ii) — Exponential form: ${\log }_{2}16=4$

Write ${\log }_{2}16=4$ in exponential form.

1. $2^4=16$.

Answer: $2^4=16$

Q4(iii) — Exponential form: ${\log }_{23}1=0$

Write ${\log }_{23}1=0$ in exponential form.

1. Any nonzero base to the power $0$ equals $1$: $23^0=1$.

Answer: $23^0=1$

Q4(iv) — Exponential form: ${\log }_{5}5=1$

Write ${\log }_{5}5=1$ in exponential form.

1. Any base to the power $1$ equals itself: $5^1=5$.

Answer: $5^1=5$

Q4(v) — Exponential form: ${\log }_2\frac{1}{8}=-3$

Write ${\log }_2\frac{1}{8}=-3$ in exponential form.

1. $2^{-3}=\frac{1}{8}$.

Answer: $2^{-3}=\frac{1}{8}$

Q4(vi) — Exponential form: $\frac{1}{2}={\log }_{9}3$

Write $\frac{1}{2}={\log }_{9}3$ in exponential form.

1. $9^{1/2}=3$.

Answer: $9^{\frac{1}{2}}=3$

Q4(vii) — Exponential form: $5={\log }_{10}100000$

Write $5={\log }_{10}100000$ in exponential form.

1. $10^5=100000$.

Answer: $10^5=100000$

Q4(viii) — Exponential form: ${\log }_4\frac{1}{16}=-2$

Write ${\log }_4\frac{1}{16}=-2$ in exponential form.

1. $4^{-2}=\frac{1}{16}$.

Answer: $4^{-2}=\frac{1}{16}$

Q4(ix) — Exponential form: ${\log }_{4}8=x$

Write ${\log }_{4}8=x$ in exponential form.

1. $4^x=8$.

Answer: $4^x=8$

Q4(x) — Exponential form: ${\log }_{9}729=3$

Write ${\log }_{9}729=3$ in exponential form.

1. $9^3=729$.

Answer: $9^3=729$

Q4(xi) — Exponential form: ${\log }_{4}1024=5$

Write ${\log }_{4}1024=5$ in exponential form.

1. $4^5=1024$.

Answer: $4^5=1024$

Q5(i) — Find $x$: ${\log }_{x}64=3$

Find $x$ if ${\log }_{x}64=3$.

1. Convert to exponential form: $x^3=64$.
2. $x=\sqrt[3]{64}=4$.

Answer: $x=4$

Q5(ii) — Find $x$: ${\log }_{5}1=x$

Find $x$ if ${\log }_{5}1=x$.

1. $\log$ of $1$ in any base is always $0$, since (base)${}^0=1$.

Answer: $x=0$

Q5(iii) — Find $x$: ${\log }_{x}8=1$

Find $x$ if ${\log }_{x}8=1$.

1. $x^1=8\Rightarrow x=8$.

Answer: $x=8$

Q5(iv) — Find $x$: ${\log }_{10}x=-3$

Find $x$ if ${\log }_{10}x=-3$.

1. $x=10^{-3}=0.001$.

Answer: $x=0.001$

Q5(v) — Find $x$: ${\log }_{4}x=\frac{3}{2}$

Find $x$ if ${\log }_{4}x=\frac{3}{2}$.

1. $x=4^{3/2}=(2^2{)}^{3/2}=2^3=8$.

Answer: $x=8$

Q5(vi) — Find $x$: ${\log }_{2}1024=x$

Find $x$ if ${\log }_{2}1024=x$.

1. $2^{10}=1024$, so $x=10$.

Answer: $x=10$

Q5(vii) — Find $x$: ${\log }_{9}x=0.5$

Find $x$ if ${\log }_{9}x=0.5$.

1. $x=9^{0.5}=\sqrt{9}=3$.

Answer: $x=3$

Q5(viii) — Find $x$: ${\left(\frac{1}{9}\right)}^{3x}=27$

Find $x$ if ${\left(\frac{1}{9}\right)}^{3x}=27$.

1. Write both sides as powers of $3$: $\frac{1}{9}=3^{-2}$, $27=3^3$.
2. $(3^{-2}{)}^{3x}=3^3\Rightarrow 3^{-6x}=3^3\Rightarrow -6x=3\Rightarrow x=-\frac{1}{2}$.

Answer: $x=-\frac{1}{2}$

Q5(ix) — Find $x$: ${\left(\frac{1}{32}\right)}^{2x}=64$

Find $x$ if ${\left(\frac{1}{32}\right)}^{2x}=64$.

1. $\frac{1}{32}=2^{-5}$, $64=2^6$.
2. $(2^{-5}{)}^{2x}=2^6\Rightarrow -10x=6\Rightarrow x=-\frac{3}{5}$.

Answer: $x=-\frac{3}{5}$

Q6(i) — Find $x$: $\log 2+\log x=1$

Find $x$ if $\log 2+\log x=1$.

1. Combine: $\log (2x)=1$.
2. $2x=10^1=10\Rightarrow x=5$.

Answer: $x=5$

Q6(ii) — Find $x$: ${\log }_{2}x+{\log }_{2}8=5$

Find $x$ if ${\log }_{2}x+{\log }_{2}8=5$.

1. Combine: ${\log }_2(8x)=5$.
2. $8x=2^5=32\Rightarrow x=4$.

Answer: $x=4$

Q6(iii) — Find $x$: $(81{)}^x=(243{)}^{x+2}$

Find $x$ if $81^x=243^{x+2}$.

1. $81=3^4$, $243=3^5$.
2. $3^{4x}=3^{5(x+2)}=3^{5x+10}$.
3. $4x=5x+10\Rightarrow x=-10$.

Answer: $x=-10$

Q6(iv) — Find $x$: ${\left(\frac{1}{27}\right)}^{x-6}=27$

Find $x$ if ${\left(\frac{1}{27}\right)}^{x-6}=27$.

1. $\frac{1}{27}=3^{-3}$, $27=3^3$.
2. $3^{-3(x-6)}=3^3\Rightarrow -3(x-6)=3\Rightarrow x-6=-1\Rightarrow x=5$.

Answer: $x=5$

Q6(v) — Find $x$: $\log (5x-10)=2$

Find $x$ if $\log (5x-10)=2$.

1. $5x-10=10^2=100$.
2. $5x=110\Rightarrow x=22$.

Answer: $x=22$

Q6(vi) — Find $x$: ${\log }_2(x+1)-{\log }_2(x-4)=2$

Find $x$ if ${\log }_2(x+1)-{\log }_2(x-4)=2$.

1. Combine: ${\log }_2\frac{x+1}{x-4}=2\Rightarrow \frac{x+1}{x-4}=4$.
2. $x+1=4x-16\Rightarrow 17=3x\Rightarrow x=\frac{17}{3}$.
3. Check domain ($x>4$): $\frac{17}{3}\approx 5.67>4$ ✓ valid.

Answer: $x=\frac{17}{3}$

Q7(i) — Characteristic of $5287$

Find the characteristic of $\log 5287$.

1. $5287$ has $4$ digits before the decimal point.
2. Characteristic $=$ (number of digits) $-1=4-1=3$.

Answer: $3$

Q7(ii) — Characteristic of $59.28$

Find the characteristic of $\log 59.28$.

1. $59.28$ has $2$ digits before the decimal point.
2. Characteristic $=2-1=1$.

Answer: $1$

Q7(iii) — Characteristic of $0.0567$

Find the characteristic of $\log 0.0567$.

1. The first significant figure ($5$) is in the $2$nd decimal place.
2. Characteristic $=-2$ (written $\bar{2}$).

Answer: $-2$ (i.e. $\bar{2}$)

Q7(iv) — Characteristic of $234.7$

Find the characteristic of $\log 234.7$.

1. $234.7$ has $3$ digits before the decimal point.
2. Characteristic $=3-1=2$.

Answer: $2$

Q7(v) — Characteristic of $0.000049$

Find the characteristic of $\log 0.000049$.

1. The first significant figure ($4$) is in the $5$th decimal place.
2. Characteristic $=-5$ (written $\bar{5}$).

Answer: $-5$ (i.e. $\bar{5}$)

Q7(vi) — Characteristic of $145000$

Find the characteristic of $\log 145000$.

1. $145000$ has $6$ digits before the decimal point.
2. Characteristic $=6-1=5$.

Answer: $5$

Q8(i) — Logarithm of $43$

Find $\log 43$ using a log table.

1. Characteristic $=1$ (2-digit number).
2. Mantissa from the log table for $4.3$ $\approx 0.6335$.

Answer: $\log 43\approx 1.6335$

Q8(ii) — Logarithm of $579$

Find $\log 579$ using a log table.

1. Characteristic $=2$ (3-digit number).
2. Mantissa for $5.79\approx 0.7627$.

Answer: $\log 579\approx 2.7627$

Q8(iii) — Logarithm of $1.982$

Find $\log 1.982$ using a log table.

1. Characteristic $=0$ (one digit before decimal).
2. Mantissa for $1.982\approx 0.2973$.

Answer: $\log 1.982\approx 0.2973$

Q8(iv) — Logarithm of $0.0876$

Find $\log 0.0876$ using a log table.

1. Characteristic $=-2$ (first significant figure in 2nd decimal place).
2. Mantissa for $8.76\approx 0.9428$.

Answer: $\log 0.0876\approx \bar{2}.9428$ (i.e. $-1.0572$)

Q8(v) — Logarithm of $0.047$

Find $\log 0.047$ using a log table.

1. Characteristic $=-2$.
2. Mantissa for $4.7\approx 0.6721$.

Answer: $\log 0.047\approx \bar{2}.6721$ (i.e. $-1.3279$)

Q8(vi) — Logarithm of $0.000354$

Find $\log 0.000354$ using a log table.

1. Characteristic $=-4$.
2. Mantissa for $3.54\approx 0.5490$.

Answer: $\log 0.000354\approx \bar{4}.5490$ (i.e. $-3.4510$)

Q9(i) — Using $\log 3.177=0.5019$, find $\log 3177$

Given $\log 3.177=0.5019$, find $\log 3177$.

1. $3177=3.177\times 10^3$, so characteristic $=3$.
2. $\log 3177=3+0.5019$.

Answer: $3.5019$

Q9(ii) — Find $\log 31.77$

Given $\log 3.177=0.5019$, find $\log 31.77$.

1. $31.77=3.177\times 10^1$, characteristic $=1$.
2. $\log 31.77=1+0.5019$.

Answer: $1.5019$

Q9(iii) — Find $\log 0.03177$

Given $\log 3.177=0.5019$, find $\log 0.03177$.

1. $0.03177=3.177\times 10^{-2}$, characteristic $=-2$.
2. $\log 0.03177=-2+0.5019$.

Answer: $\bar{2}.5019$ (i.e. $-1.4981$)

Q10(i) — Find $x$: $\log x=0.0065$

Find $x$ if $\log x=0.0065$ (antilog).

1. $x=10^{0.0065}$; since the mantissa is close to $0$, $x$ is close to $1$.

Answer: $x\approx 1.015$

Q10(ii) — Find $x$: $\log x=1.192$

Find $x$ if $\log x=1.192$.

1. Characteristic $1$ means $x$ has $2$ digits before the decimal.
2. Antilog of mantissa $0.192\approx 1.556$, so $x\approx 15.56$.

Answer: $x\approx 15.56$

Q10(iii) — Find $x$: $\log x=-3.434$

Find $x$ if $\log x=-3.434$.

1. Rewrite with a positive mantissa: $-3.434=-4+0.566=\bar{4}.566$.
2. Antilog of mantissa $0.566\approx 3.681$, and characteristic $-4$ shifts the decimal: $x\approx 3.681\times 10^{-4}$.

Answer: $x\approx 0.0003681$

Q10(iv) — Find $x$: $\log x=-1.5726$

Find $x$ if $\log x=-1.5726$.

1. Rewrite: $-1.5726=-2+0.4274=\bar{2}.4274$.
2. Antilog of $0.4274\approx 2.676$, so $x\approx 2.676\times 10^{-2}$.

Answer: $x\approx 0.02676$

Q10(v) — Find $x$: $\log x=4.3561$

Find $x$ if $\log x=4.3561$.

1. Characteristic $4$ means $x$ has $5$ digits before the decimal.
2. Antilog of $0.3561\approx 2.271$, so $x\approx 2.271\times 10^4$.

Answer: $x\approx 22710$

Q10(vi) — Find $x$: $\log x=-2.0184$

Find $x$ if $\log x=-2.0184$.

1. Rewrite: $-2.0184=-3+0.9816=\bar{3}.9816$.
2. Antilog of $0.9816\approx 9.585$, so $x\approx 9.585\times 10^{-3}$.

Answer: $x\approx 0.009585$

Q11(i) — Evaluate: ${\log }_{2}18-{\log }_{2}9$

Without a calculator, evaluate ${\log }_{2}18-{\log }_{2}9$.

1. Quotient law: ${\log }_2\frac{18}{9}={\log }_{2}2$.

Answer: $1$

Q11(ii) — Evaluate: ${\log }_{2}64+{\log }_{2}2$

Evaluate ${\log }_{2}64+{\log }_{2}2$.

1. Product law: ${\log }_2(64\times 2)={\log }_{2}128$.
2. $2^7=128$.

Answer: $7$

Q11(iii) — Evaluate: $\frac{1}{3}{\log }_{3}8-{\log }_{3}18$

Evaluate $\frac{1}{3}{\log }_{3}8-{\log }_{3}18$.

1. $\frac{1}{3}{\log }_{3}8={\log }_3{8}^{1/3}={\log }_{3}2$.
2. ${\log }_{3}2-{\log }_{3}18={\log }_3\frac{2}{18}={\log }_3\frac{1}{9}={\log }_3{3}^{-2}=-2$.

Answer: $-2$

Q11(iv) — Evaluate: $2\log 2+\log 25$

Evaluate $2\log 2+\log 25$.

1. $2\log 2=\log 4$.
2. $\log 4+\log 25=\log (4\times 25)=\log 100=2$.

Answer: $2$

Q11(v) — Evaluate: $\frac{1}{3}{\log }_{4}64+2{\log }_{5}25$

Evaluate $\frac{1}{3}{\log }_{4}64+2{\log }_{5}25$.

1. ${\log }_{4}64={\log }_4{4}^3=3$, so $\frac{1}{3}(3)=1$.
2. ${\log }_{5}25={\log }_5{5}^2=2$, so $2(2)=4$.
3. Sum $=1+4=5$.

Answer: $5$

Q11(vi) — Evaluate: ${\log }_{3}12+{\log }_{3}0.25$

Evaluate ${\log }_{3}12+{\log }_{3}0.25$.

1. Product law: ${\log }_3(12\times 0.25)={\log }_{3}3=1$.

Answer: $1$

Q12(i) — Single logarithm: $\frac{1}{2}\log 25+2\log 3$

Write $\frac{1}{2}\log 25+2\log 3$ as a single logarithm.

1. $\frac{1}{2}\log 25=\log 25^{1/2}=\log 5$.
2. $2\log 3=\log 9$.
3. $\log 5+\log 9=\log (5\times 9)=\log 45$.

Answer: $\log 45$

Q12(ii) — Single logarithm: $\log 9-\log \frac{1}{3}$

Write $\log 9-\log \frac{1}{3}$ as a single logarithm.

1. $\log \left(9÷\frac{1}{3}\right)=\log (9\times 3)=\log 27$.

Answer: $\log 27$

Q12(iii) — Single logarithm: ${\log }_5{b}^2\cdot {\log }_a{5}^3$

Write ${\log }_5{b}^2\cdot {\log }_a{5}^3$ as a single logarithm.

1. $=2{\log }_{5}b\times 3{\log }_{a}5=6({\log }_{5}b\times {\log }_{a}5)$.
2. By the chain rule for logs (via a common base), ${\log }_{5}b\times {\log }_{a}5={\log }_{a}b$.
3. So the product is $6{\log }_{a}b={\log }_a{b}^6$.

Answer: ${\log }_a{b}^6$

Q12(iv) — Single logarithm: $2{\log }_{3}x+{\log }_{3}y$

Write $2{\log }_{3}x+{\log }_{3}y$ as a single logarithm.

1. $2{\log }_{3}x={\log }_3{x}^2$.
2. ${\log }_3{x}^2+{\log }_{3}y={\log }_3(x^{2}y)$.

Answer: ${\log }_3(x^{2}y)$

Q12(v) — Single logarithm: $4{\log }_{5}x-{\log }_{5}y+{\log }_{5}z$

Write $4{\log }_{5}x-{\log }_{5}y+{\log }_{5}z$ as a single logarithm.

1. $4{\log }_{5}x={\log }_5{x}^4$.
2. Combine: ${\log }_5\frac{x^{4}z}{y}$.

Answer: ${\log }_5\left(\frac{x^{4}z}{y}\right)$

Q12(vi) — Single logarithm: $2\ln a+3\ln b-4\ln c$

Write $2\ln a+3\ln b-4\ln c$ as a single logarithm.

1. $=\ln a^2+\ln b^3-\ln c^4=\ln \frac{a^2{b}^3}{c^4}$.

Answer: $\ln \left(\frac{a^2{b}^3}{c^4}\right)$

Q12(vii) — Single logarithm: $7\log x-3\log y^2$

Write $7\log x-3\log y^2$ as a single logarithm.

1. $3\log y^2=\log (y^2{)}^3=\log y^6$.
2. $\log x^7-\log y^6=\log \frac{x^7}{y^6}$.

Answer: $\log \left(\frac{x^7}{y^6}\right)$

Q12(viii) — Single logarithm: $3\log 4-\log 32$

Write $3\log 4-\log 32$ as a single logarithm.

1. $3\log 4=\log 4^3=\log 64$.
2. $\log 64-\log 32=\log \frac{64}{32}=\log 2$.

Answer: $\log 2$

Q12(ix) — Single logarithm: $\frac{1}{3}({\log }_{5}8+{\log }_{5}27)-{\log }_{5}3$

Write $\frac{1}{3}({\log }_{5}8+{\log }_{5}27)-{\log }_{5}3$ as a single logarithm.

1. ${\log }_{5}8+{\log }_{5}27={\log }_5(8\times 27)={\log }_{5}216$.
2. $216=6^3$, so $\frac{1}{3}{\log }_{5}216={\log }_{5}6$.
3. ${\log }_{5}6-{\log }_{5}3={\log }_5\frac{6}{3}={\log }_{5}2$.

Answer: ${\log }_{5}2$

Q13(i) — Expand: $\log \left(\frac{11}{5}\right)$

Expand $\log \left(\frac{11}{5}\right)$ using the laws of logarithms.

1. Quotient law: $\log 11-\log 5$.

Answer: $\log 11-\log 5$

Q13(ii) — Expand: ${\log }_5\sqrt{8a^6}$

Expand ${\log }_5\sqrt{8a^6}$.

1. $\sqrt{8a^6}=(8a^6{)}^{1/2}$, so ${\log }_5(8a^6{)}^{1/2}=\frac{1}{2}{\log }_5(8a^6)$.
2. $=\frac{1}{2}({\log }_{5}8+{\log }_5{a}^6)=\frac{1}{2}{\log }_{5}8+3{\log }_{5}a$.

Answer: $\frac{1}{2}{\log }_{5}8+3{\log }_{5}a$

Q13(iii) — Expand: $\ln \left(\frac{a^{2}b}{c}\right)$

Expand $\ln \left(\frac{a^{2}b}{c}\right)$.

1. $=\ln a^2+\ln b-\ln c=2\ln a+\ln b-\ln c$.

Answer: $2\ln a+\ln b-\ln c$

Q13(iv) — Expand: $\log {\left(\frac{xy}{z}\right)}^{\frac{1}{9}}$

Expand $\log {\left(\frac{xy}{z}\right)}^{1/9}$.

1. $=\frac{1}{9}\log \left(\frac{xy}{z}\right)=\frac{1}{9}(\log x+\log y-\log z)$.

Answer: $\frac{1}{9}(\log x+\log y-\log z)$

Q13(v) — Expand: $\ln \sqrt[3]{16x^3}$

Expand $\ln \sqrt[3]{16x^3}$.

1. $=\frac{1}{3}\ln (16x^3)=\frac{1}{3}(\ln 16+3\ln x)=\frac{1}{3}\ln 16+\ln x$.
2. Since $16=2^4$, $\ln 16=4\ln 2$, so this also equals $\frac{4}{3}\ln 2+\ln x$.

Answer: $\frac{1}{3}\ln 16+\ln x$ (equivalently $\frac{4}{3}\ln 2+\ln x$)

Q13(vi) — Expand: ${\log }_2{\left(\frac{1-a}{b}\right)}^5$

Expand ${\log }_2{\left(\frac{1-a}{b}\right)}^5$.

1. $=5{\log }_2\left(\frac{1-a}{b}\right)=5[{\log }_2(1-a)-{\log }_{2}b]$.

Answer: $5{\log }_2(1-a)-5{\log }_{2}b$

Q13(vii) — Expand: $\log (xy{z}^6)$

Expand $\log (xy{z}^6)$.

1. Product law across all three factors: $\log x+\log y+\log z^6=\log x+\log y+6\log z$.

Answer: $\log x+\log y+6\log z$

Q13(viii) — Expand: ${\log }_3\sqrt[6]{m^5{n}^3}$

Expand ${\log }_3\sqrt[6]{m^5{n}^3}$.

1. $=\frac{1}{6}{\log }_3(m^5{n}^3)=\frac{1}{6}(5{\log }_{3}m+3{\log }_{3}n)=\frac{5}{6}{\log }_{3}m+\frac{1}{2}{\log }_{3}n$.

Answer: $\frac{5}{6}{\log }_{3}m+\frac{1}{2}{\log }_{3}n$

Q13(ix) — Expand: $\log \sqrt{8x^3}$

Expand $\log \sqrt{8x^3}$.

1. $=\frac{1}{2}\log (8x^3)=\frac{1}{2}(\log 8+3\log x)$.
2. $\log 8=3\log 2$, so $=\frac{3}{2}\log 2+\frac{3}{2}\log x$.

Answer: $\frac{3}{2}\log 2+\frac{3}{2}\log x$

Q14(i) — $\frac{3.68\times 4.21}{5.234}$ using logs

Evaluate $\frac{3.68\times 4.21}{5.234}$ using a log table.

1. $\log 3.68\approx 0.5659$, $\log 4.21\approx 0.6243$, $\log 5.234\approx 0.7188$.
2. Result's $\log =0.5659+0.6243-0.7188=0.4714$.
3. Antilog of $0.4714\approx 2.961$.

Answer: $\approx 2.961$

Q14(ii) — $4.67\times 2.11\times 2.397$ using logs

Evaluate $4.67\times 2.11\times 2.397$ using a log table.

1. $\log 4.67\approx 0.6694$, $\log 2.11\approx 0.3243$, $\log 2.397\approx 0.3794$.
2. Sum $=1.3731$.
3. Antilog of $1.3731\approx 23.61$.