Class 11- Applied Mathematics-NCERT Handbook Solutions-Chapter-1

Solution:

To convert a binary number to a decimal number, we expand it by multiplying each binary digit by its corresponding power of $2$, starting from $2^0$ on the far right.

1. ${\left(1010101100110\right)}_2=1\cdot 2^{12}+0\cdot 2^{11}+1\cdot 2^{10}+0\cdot 2^9+1\cdot 2^8+0\cdot 2^7+1\cdot 2^6+1\cdot 2^5+0\cdot 2^4+0\cdot 2^3+1\cdot 2^2+1\cdot 2^1+0\cdot 2^0$
   
   $=4096+0+1024+0+256+0+64+32+0+0+4+2+0$
   
   $=5478$


2. ${\left(101011000110\right)}_2=1\cdot 2^{11}+0\cdot 2^{10}+1\cdot 2^9+0\cdot 2^8+1\cdot 2^7+1\cdot 2^6+0\cdot 2^5+0\cdot 2^4+0\cdot 2^3+1\cdot 2^2+1\cdot 2^1+0\cdot 2^0$
   
   $=2048+0+512+0+128+64+0+0+0+4+2+0$
   
   $=2758$


3. ${\left(101111100110\right)}_2=1\cdot 2^{11}+0\cdot 2^{10}+1\cdot 2^9+1\cdot 2^8+1\cdot 2^7+1\cdot 2^6+1\cdot 2^5+0\cdot 2^4+0\cdot 2^3+1\cdot 2^2+1\cdot 2^1+0\cdot 2^0$
   
   $=2048+0+512+256+128+64+32+0+0+4+2+0$
   
   $=3046$


4. ${\left(1000000000110\right)}_2=1\cdot 2^{12}+0\cdot 2^{11}+0\cdot 2^{10}+0\cdot 2^9+0\cdot 2^8+0\cdot 2^7+0\cdot 2^6+0\cdot 2^5+0\cdot 2^4+0\cdot 2^3+1\cdot 2^2+1\cdot 2^1+0\cdot 2^0$
   
   $=4096+0+0+0+0+0+0+0+0+0+4+2+0$
   
   $=4102$

Solution:

Note: The problem statement asks to write the numbers in decimal notation, but the provided numbers are already in decimal format. Based on the textbook's answer key, the intended task is to convert these decimal numbers into binary notation. We will proceed to convert them to binary. For the last number, the answer key provides the binary evaluation for $2468097531$ rather than $246809753$. We will provide the solution for $2468097531$ to reflect the correct key.

To convert a decimal number to binary, repeatedly divide the number by $2$ and record the remainders. The binary representation is formed by reading the remainders from the last division up to the first.

• By repeatedly dividing $654321$ by $2$, we get:

  ${\left(654321\right)}_{10}={\left(1001111110111111110001\right)}_2$

• By repeatedly dividing $1000001$ by $2$, we get:

  ${\left(1000001\right)}_{10}={\left(11110100001001000001\right)}_2$

• By repeatedly dividing $56237801$ by $2$, we get:

  ${\left(56237801\right)}_{10}={\left(110101101010000111101110101001\right)}_2$

• By repeatedly dividing $2468097531$ by $2$, we get:

  ${\left(2468097531\right)}_{10}={\left(10010011010011010101111101111011\right)}_2$

Solution:

Using the product law of exponents: $x^m\cdot x^n=x^{m+n}$

$a=x^{\frac{1}{5}+\frac{4}{5}}=x^{\frac{1+4}{5}}=x^{\frac{5}{5}}=x^1=x$

Solution:

Using the quotient law of exponents: $\frac{x^m}{x^n}=x^{m-n}$

$a=x^{2-\frac{3}{2}}=x^{\frac{4-3}{2}}=x^{\frac{1}{2}}$

Solution:

Note: The textbook exhibits a typographical error showing the fraction as $\frac{3}{3}$, but based on the provided answer key making the result $x^{10}$, the intended fraction is $\frac{5}{3}$.

Using the power of a power law of exponents: ${\left(x^m\right)}^n=x^{m\cdot n}$

$a=x^{\frac{5}{3}\cdot 6}=x^{5\cdot 2}=x^{10}$

Solution:

Using the quotient law of exponents: $\frac{x^m}{x^n}=x^{m-n}$

$a=x^{3-5}\cdot y^{4-2}=x^{-2}{y}^2=\frac{y^2}{x^2}$

Solution:

(i) We can express 15 as the product of its prime factors:

$\log 15=\log (3\times 5)$

Using the product law of logarithms $\log \left(mn\right)=\log m+\log n$:

$\log 15=\log 3+\log 5$ (Proved)

(ii) We find the prime factorization of 500:

$500=4\times 125=2^2\times 5^3$

Applying the product and power laws of logarithms:

$\log 500=\log (2^2\times 5^3)$

$=\log \left(2^2\right)+\log \left(5^3\right)$

$=2\log 2+3\log 5$ (Proved)

(iii) Extending the product law to three terms $\log \left(abc\right)=\log a+\log b+\log c$:

$\log (1\times 2\times 3)=\log 1+\log 2+\log 3$ (Proved)

(iv) Using the product law in reverse to combine the logarithms:

$\mathrm{LHS}=\log (\frac{x^2}{yz}\times \frac{y^2}{zx}\times \frac{z^2}{xy})$

$=\log \left(\frac{x^2{y}^2{z}^2}{x^2{y}^2{z}^2}\right)$

$=\log \left(1\right)=0=\mathrm{RHS}$ (Proved)

Solution:

Applying the quotient, product, and power rules of logarithms:

${\log }_b\left(\frac{a^2{b}^b}{c^c{d}^d}\right)={\log }_b\left(a^2{b}^b\right)-{\log }_b\left(c^c{d}^d\right)$

$=[{\log }_b(a^2)+{\log }_b(b^b\left)\right]-[{\log }_b(c^c)+{\log }_b(d^d\left)\right]$

$=2{\log }_{b}a+b{\log }_{b}b-c{\log }_{b}c-d{\log }_{b}d$

Since ${\log }_{b}b=1$, this simplifies to:

$=2{\log }_{b}a+b-c{\log }_{b}c-d{\log }_{b}d$

Solution:

Applying the logarithm laws systematically:

${\log }_b\left(\frac{4x^6}{9y^7}\right)={\log }_b\left(4x^6\right)-{\log }_b\left(9y^7\right)$

$=[{\log }_b(2^2)+{\log }_b(x^6\left)\right]-[{\log }_b(3^2)+{\log }_b(y^7\left)\right]$

$=2{\log }_{b}2+6{\log }_{b}x-2{\log }_{b}3-7{\log }_{b}y$

Solution:

Using the product law to combine terms with the same base:

${\log }_{10}a+{\log }_{10}{b}^2+{\log }_{10}{c}^3={\log }_{10}(a\cdot b^2\cdot c^3)$

Solution:

First apply the power rule, then substitute ${\log }_{x}x=1$ for each term:

$=1-2{\log }_{b}b+3{\log }_{c}c-4{\log }_{d}d$

$=1-2\left(1\right)+3\left(1\right)-4\left(1\right)$

$=1-2+3-4=-2$

Solution:

(i) Assuming common logarithms (base 10), we can write the constant $2$ as $\log 100$:

$=\frac{3}{2}\log \left(2^2\right)+\frac{1}{3}\log \left(3^3\right)+\log \left(5^2\right)-\log 6-\log 100$

$=3\log 2+\log 3+\log 25-\log 6-\log 100$

$=\log \left(2^3\right)+\log 3+\log 25-(\log 6+\log 100)$

$=\log (8\times 3\times 25)-\log (6\times 100)$

$=\log \left(600\right)-\log \left(600\right)=0$

(ii) Using the product law:

$\log 25+\log 200=\log (25\times 200)$

$=\log 5000=\log (5\times {10}^3)$

$=\log 5+\log {10}^3=\log 5+3$

(iii) Using the quotient law and algebraic factoring:

${\log }_5(x^2-1)-{\log }_5(x-1)={\log }_5\left(\frac{x^2-1}{x-1}\right)$

$={\log }_5\left(\frac{(x-1)(x+1)}{x-1}\right)$

$={\log }_5(x+1)$

Solution:

A logarithm written with a bar over the characteristic (integer part), such as $\overline{1}.4914$, indicates that only the integer part is negative, while the decimal part (mantissa) is positive. Thus:

$\log x=-1+0.4914=-0.5086$
$\log y=2.4669$

(i) Find $xy$:

$\log \left(xy\right)=\log x+\log y$

$=-0.5086+2.4669=1.9583$

Now, calculate the antilog of 1.9583:

$xy=\mathrm{Antilog}\left(1.9583\right)\approx 90.84$

Note: If $\log x$ was intended to be given as $1.4914$ without the bar, the logarithm sum would be $3.9583$.

(ii) Find $\frac{x}{y}$:

$\log \left(\frac{x}{y}\right)=\log x-\log y$

$=-0.5086-2.4669=-2.9755$

To find the antilog, we must convert this into a format with a positive mantissa:

$-2.9755=-3+0.0245=\overline{3}.0245$

$\frac{x}{y}=\mathrm{Antilog}(\overline{3}.0245)\approx 0.001058$

Note: If $\log x$ was intended to be given as positive $1.4914$, $\log (x/y)=\overline{1}.0245$, yielding the answer $0.106$.

(iii) Find $\frac{x^3}{y^2}$:

$\log \left(\frac{x^3}{y^2}\right)=3\log x-2\log y$

$=3(-0.5086)-2\left(2.4669\right)$

$=-1.5258-4.9338=-6.4596$

Again, converting to a format with a positive mantissa:

$-6.4596=-7+0.5404=\overline{7}.5404$

$\frac{x^3}{y^2}=\mathrm{Antilog}(\overline{7}.5404)\approx 3.47\times {10}^{-7}$

Note: If $\log x$ was intended to be given as positive $1.4914$, $\log (x^3/y^2)=\overline{1}.5404$, yielding the answer $0.347$.

Solution:

First, apply the power rule of logarithms to find ${\log }_{10}\left(2^{32}\right)$:

$\log \left(2^{32}\right)=32\log 2$

Using the standard logarithm value $\log 2\approx 0.30103$:

$\log \left(2^{32}\right)\approx 32\times 0.30103=9.63296$

The integer part (characteristic) of the logarithm is $9$.

The number of digits in a whole number is calculated by adding 1 to the characteristic of its base-10 logarithm:

$\text{Number of digits}=9+1=10$

Therefore, $2^{32}$ has 10 digits.

Solution:

Using the Aryabhatan system of numeration, we split words into sub-units of consonants and vowels, assigning values based on the Varga/Avarga tables and multiplying the consonant value by the vowel's power of $10$.

1. krșņā (कृष्णा):

   Split into (k + ṛ) and (ș + ņ + ā)

   $=(1\times {10}^6)+(80+15)\times {10}^0$
   $=1000000+95=1000095$

2. mukti (मुक्ति):

   Split into (m + u) and (k + t + i)

   $=(25\times {10}^4)+(1+16)\times {10}^2$
   $=250000+1700=251700$

3. jyestha (ज्येष्ठ):

   Split into (j + y + e) and (ș + ṭ + a)

   $=(8+30)\times {10}^{10}+(80+12)\times {10}^0$
   $=38\times {10}^{10}+92=380000000092$

Solution:

The number is $3\times {10}^8$. According to the Aryabhatan tables:

• The consonant for $3$ is 'ga' (ग).
• The vowel for ${10}^8$ is 'ḷ' (ऌ).

Combining them, we get: ग + ऌ = गॣ (gḷ)

Solution: (b) 9

${6561}^{0.14+0.11}={6561}^{0.25}={\left(3^8\right)}^{\frac{1}{4}}=3^2=9$

Solution: (a) Not Defined

Logarithms are only defined for positive real numbers. Since the argument is $0$, it is undefined.

Solution: (a) 64

$3^{\frac{n}{2}}=3^8\Rightarrow \frac{n}{2}=8\Rightarrow n=16$

${\left(16\right)}^{\frac{3}{2}}={\left(4^2\right)}^{\frac{3}{2}}=4^3=64$

Solution: (d) 22.5

$5^x\times 8=36\Rightarrow 5^x=\frac{36}{8}=4.5$

$5^{(x+1)}=5^x\times 5=4.5\times 5=22.5$

Solution: (d) None / Out of options

Note: Based on the options provided in the textbook (44, 44, 62, 64) and answer key, if the question was formatted as $8^{120}={(8\times x)}^{40}$, solving gives $x=64$ which aligns with the answer key (d). For the literal text:

${\left({36}^3\right)}^{40}={\left(36x\right)}^{40}\Rightarrow {36}^3=36x\Rightarrow x={36}^2=1296$

Solution: (c) 93

$={\left(3^8\right)}^{\frac{1}{2}}+{\left(3^8\right)}^{\frac{1}{4}}+{\left(3^8\right)}^{\frac{1}{8}}=3^4+3^2+3^1$
$=81+9+3=93$

Solution: (c)

The addition of numbers within a log argument does not translate to the product of numbers. Specifically, $\log (3+4)=\log \left(7\right)$, which is not equal to $\log \left(12\right)$.

Solution: (d) 36

$5^{(a+b)}=5^1\times 5^2\times 5^3=5^{1+2+3}=5^6$
$a+b=6\Rightarrow {(a+b)}^2=6^2=36$

Solution: (a) $6\times 7^{-15}$

$7^{-14}-7^{-15}=7^{-15}(7^1-1)=7^{-15}\left(6\right)=6\times 7^{-15}$

Solution: (c) -2

Factor out the common term:

$3^{n+2}(3^2-1)=8$
$3^{n+2}\left(8\right)=8\Rightarrow 3^{n+2}=1$
$3^{n+2}=3^0\Rightarrow n+2=0\Rightarrow n=-2$

Solution: (c) 10/7

$2^x={\left(1024\right)}^{\frac{1}{7}}={\left(2^{10}\right)}^{\frac{1}{7}}=2^{\frac{10}{7}}\Rightarrow x=\frac{10}{7}$

Solution: (a) 15/4

${\left[{(3^{-2}-5^{-2})}^{17-18}\right]}^{\frac{1}{2}}={\left[{(\frac{1}{9}-\frac{1}{25})}^{-1}\right]}^{\frac{1}{2}}$
$={\left[{\left(\frac{25-9}{225}\right)}^{-1}\right]}^{\frac{1}{2}}={\left[{\left(\frac{16}{225}\right)}^{-1}\right]}^{\frac{1}{2}}$
$={\left[\frac{225}{16}\right]}^{\frac{1}{2}}=\sqrt{\frac{225}{16}}=\frac{15}{4}$

Solution:

We convert decimal to binary by successively dividing by 2 and reading the remainders upwards.

1. $4={\left(100\right)}_2$
2. $49=32+16+1={\left(110001\right)}_2$
3. $267=256+8+2+1={\left(100001011\right)}_2$
4. $1024=2^{10}={\left(10000000000\right)}_2$
5. $4103=4096+4+2+1={\left(1000000000111\right)}_2$

Solution:

1. ${\left(1000\right)}_2=1\cdot 2^3=8$
2. ${\left(11001\right)}_2=16+8+1=25$
3. ${\left(1110101\right)}_2=64+32+16+4+1=117$
4. ${\left(10011001\right)}_2=128+16+8+1=153$

Solution:

Using the property $b^x=y\Rightarrow {\log }_{b}y=x$:

1. ${\log }_{10}1000=3$
2. ${\log }_{10}\left(0.01\right)=-2$
3. ${\log }_{0.2}\left(0.008\right)=3$
4. ${\log }_4\left(\frac{1}{16}\right)=-2$
5. ${\log }_{81}27=\frac{3}{4}$

Solution:

1. $2^1=2$
2. $5^{-1}=\frac{1}{5}$
3. ${10}^{-3}=0.001$
4. ${27}^{\frac{2}{3}}=9$
5. $4^0=1$

Solution:

1. ${\log }_{3}x=2\Rightarrow x=3^2=9$
2. $x+1={\left(0.2\right)}^{-2}={\left(\frac{1}{5}\right)}^{-2}=5^2=25\Rightarrow x=24$
3. $x^2-1=2^3=8\Rightarrow x^2=9\Rightarrow x=\pm 3$
4. $x-10={10}^1=10\Rightarrow x=20$

Solution:

The characteristic is the integer part of the common logarithm, which represents the power of 10 in scientific notation.

1. $23.84=2.384\times {10}^1\Rightarrow \text{Characteristic}=1$
2. $384.76=3.8476\times {10}^2\Rightarrow \text{Characteristic}=2$
3. $9.857=9.857\times {10}^0\Rightarrow \text{Characteristic}=0$
4. $0.00035=3.5\times {10}^{-4}\Rightarrow \text{Characteristic}=-4$ (or $\overline{4}$)
5. $0.00002356=2.356\times {10}^{-5}\Rightarrow \text{Characteristic}=-5$ (or $\overline{5}$)

Solution:

The mantissa is found by reading the log tables corresponding to the significant digits.

1. For $24.6$, look up 24 under 6: Mantissa $\approx 0.3909$
2. For $348.56$, look up 34 under 8 (mean diff 6): Mantissa $\approx 0.5423$
3. For $2.768$, look up 27 under 6 (mean diff 8): Mantissa $\approx 0.4422$
4. For $0.0056$, look up 56 under 0: Mantissa $\approx 0.7482$

Solution:

Combining the characteristic and mantissa:

1. $\log 5=0.6990$
2. $\log 67=1.8261$
3. $\log 734.5=2.8660$
4. $\log \left(0.0056\right)=\overline{3}.7482=-2.2518$

Solution:

1. $x=\mathrm{Antilog}\left(0.5647\right)\times {10}^1\approx 3.670\times 10=36.70$
2. $x=\mathrm{Antilog}\left(0.9031\right)\times {10}^2\approx 8.00\times 100=800.0$
3. $x=\mathrm{Antilog}\left(0.4567\right)\times {10}^2\approx 2.862\times 100=286.2$
4. First write with a positive mantissa: $-2.1304=-3+0.8696=\overline{3}.8696$.
   $x=\mathrm{Antilog}\left(0.8696\right)\times {10}^{-3}\approx 7.406\times {10}^{-3}=0.007406$

Solution:

Flip the fraction on the right side to match the base:

${\left(\frac{a}{b}\right)}^{x-2}={\left(\frac{a}{b}\right)}^{-(x-7)}$

Equate the exponents:

$x-2=-x+7$
$2x=9\Rightarrow x=4.5$

Solution:

Express the numbers as powers of 7:

$7^{x-y}=7^3\Rightarrow x-y=3$
$7^{x+y}=7^5\Rightarrow x+y=5$

Adding both equations gives: $2x=8\Rightarrow x=4$

Solution:

(a)

${\left(0.04\right)}^{-2.5}={\left(\frac{4}{100}\right)}^{-\frac{5}{2}}={\left(\frac{1}{25}\right)}^{-\frac{5}{2}}$
$={\left(25\right)}^{\frac{5}{2}}={\left(5^2\right)}^{\frac{5}{2}}=5^5=3125$

(b)

$6^{6.5}\times {\left(6^2\right)}^{4.5}÷{\left(6^3\right)}^{4.5}=6^{6.5}\times 6^9÷6^{13.5}$
$=6^{6.5+9-13.5}=6^2$

The missing exponent is $2$.

Solution:

We know ${14}^2=196$. So, $m=14$ and $n=2$.

${(m-3)}^{n+1}={(14-3)}^{2+1}={11}^3=1331$

Solution:

(a) Bring the coefficient inside as an exponent and combine using product/quotient laws:

$=\log \left(\frac{75}{16}\right)-\log {\left(\frac{5}{9}\right)}^2+\log \left(\frac{32}{243}\right)$
$=\log (\frac{75}{16}\times \frac{81}{25}\times \frac{32}{243})$
$=\log \left(\frac{3\times 81\times 2}{243}\right)=\log \left(\frac{486}{243}\right)=\log 2$

(b)

$=\log {\left(\frac{15}{18}\right)}^2-\log \left(\frac{25}{162}\right)+\log \left(\frac{4}{9}\right)$
$=\log (\frac{225}{324}\times \frac{162}{25}\times \frac{4}{9})$
$=\log (9\times \frac{1}{2}\times \frac{4}{9})=\log \left(2\right)$

Solution:

$\log \left(\frac{10x}{y^2}\right)=\log 10+\log x-2\log y$
$=1+(m+n)-2(m-n)$
$=1+m+n-2m+2n=1-m+3n$

Solution:

Change all bases to 2 using the rule ${\log }_{a^k}b=\frac{1}{k}{\log }_{a}b$:

${\log }_{2}x+\frac{1}{2}{\log }_{2}x+\frac{1}{4}{\log }_{2}x=\frac{21}{4}$
${\log }_{2}x(1+\frac{1}{2}+\frac{1}{4})=\frac{21}{4}$
${\log }_{2}x\left(\frac{7}{4}\right)=\frac{21}{4}$
${\log }_{2}x=3\Rightarrow x=2^3=8$

Solution:

(a) Simplify terms using exponents laws:

$={\left(x^{q-r}\right)}^{q+r-p}\cdot {\left(x^{r-p}\right)}^{r+p-q}\cdot {\left(x^{p-q}\right)}^{p+q-r}$
$=x^{(q-r)(q+r-p)+(r-p)(r+p-q)+(p-q)(p+q-r)}$

Expanding the exponent sum:

$(q^2-r^2-pq+pr)+(r^2-p^2-qr+pq)+(p^2-q^2-pr+qr)=0$

$=x^0=1$

(b) Multiply the numerator and denominator of the first term by $x^p$, the second by $x^q$, and the third by $x^r$:

$=\frac{x^p}{x^p+x^q+x^r}+\frac{x^q}{x^q+x^p+x^r}+\frac{x^r}{x^r+x^p+x^q}$
$=\frac{x^p+x^q+x^r}{x^p+x^q+x^r}=1$

Solution:

Let $2^x=3^y={12}^z=k$.

Then, $2=k^{\frac{1}{x}}$, $3=k^{\frac{1}{y}}$, and $12=k^{\frac{1}{z}}$.

Since $12=2^2\cdot 3$, substituting the values gives:

$k^{\frac{1}{z}}={\left(k^{\frac{1}{x}}\right)}^2\cdot k^{\frac{1}{y}}$
$k^{\frac{1}{z}}=k^{\frac{2}{x}+\frac{1}{y}}$

Equating the exponents:

$\frac{1}{z}=\frac{2}{x}+\frac{1}{y}$
$\frac{2}{x}=\frac{1}{z}-\frac{1}{y}=\frac{y-z}{yz}$
$x=\frac{2yz}{y-z}$ (Proved)

Solution:

Express all bases as powers of 4: $1024=4^5$ and $16=4^2$.

$=\frac{{\left(4^5\right)}^{\frac{n}{5}}\times 4^{2n+1}}{{\left(4^2\right)}^n\times 4^{n-1}}$
$=\frac{4^n\times 4^{2n+1}}{4^{2n}\times 4^{n-1}}$
$=\frac{4^{3n+1}}{4^{3n-1}}$
$=4^{(3n+1)-(3n-1)}=4^2=16$

Solution:

Since multiplication is commutative, $ab=ba$. Therefore, the equation is ${\left(ab\right)}^{x-4}={\left(ab\right)}^{x-8}$.

If the base $ab$ is neither $0$ nor $1$, equality of exponents gives $x-4=x-8$, which is impossible. Thus, no general solution exists unless $ab=1$ (where all x work).

Solution:

First, simplify $x$:

$x={\left({10}^2\right)}^{0.20}={10}^{0.40}$

Now, substitute $x$ and $y$ into the equation $x^z=y^2$:

${\left({10}^{0.40}\right)}^z={\left({10}^{0.60}\right)}^2$
${10}^{0.4z}={10}^{1.2}$

Equating the exponents:

$0.4z=1.2\Rightarrow z=\frac{1.2}{0.4}=3$

Note: The textbook answer key states 4, which would correspond to an original value of $x={1000}^{0.10}$ or ${10}^{0.30}$. Mathematically, for the provided text, $z=3$.

Solution:

(i) Let $x$ equal the expression. Taking logs:

$\log x=3\log \left(2.463\right)+\frac{1}{2}\log \left(72.42\right)-\frac{1}{3}\log \left(34.59\right)$
$=3\left(0.3915\right)+0.5\left(1.8599\right)-\frac{1}{3}\left(1.5390\right)$
$=1.1745+0.92995-0.5130=1.59145$

$x=\mathrm{Antilog}\left(1.59145\right)\approx 39.04$

(ii) Let $y$ equal the expression. Taking logs:

$\log y=\frac{1}{2}[3\log (238.2)+\log (86.34)-\log (92.3\left)\right]$
$=\frac{1}{2}\left[3\right(2.3770)+1.9362-1.9652]$
$=\frac{1}{2}[7.1310-0.0290]=\frac{1}{2}\left[7.1020\right]=3.5510$

$y=\mathrm{Antilog}\left(3.5510\right)\approx 3556$

Solution:

(i) For the year 2029, $t=2029-2010=19$.

$P=4.7{\left(1.02\right)}^{19}$

Applying logarithms:

$\log P=\log \left(4.7\right)+19\log \left(1.02\right)$
$\log P\approx 0.6721+19\left(0.0086\right)=0.6721+0.1634=0.8355$

Taking the antilog:

$P=\mathrm{Antilog}\left(0.8355\right)\approx 6.847$

The population is approximately $6.847$ billion, which is $6847$ million.

(ii) Population in 2020 ($t=10$) is $P_{2020}=4.7{\left(1.02\right)}^{10}$.

We want to find $t$ such that $P=2\times P_{2020}$:

$4.7{\left(1.02\right)}^t=2\times 4.7{\left(1.02\right)}^{10}$
${\left(1.02\right)}^{t-10}=2$

Applying logarithms to both sides:

$(t-10)\log \left(1.02\right)=\log 2$
$(t-10)\left(0.0086\right)\approx 0.3010$
$t-10\approx \frac{0.3010}{0.0086}=35$
$t=10+35=45$

The year is $2010+45=2055$.

Solution: (d)

Assertion (A) is false because the domains must match. $\log (x-3)$ and $\log (x+3)$ are only defined when $x>3$, not for all $x\ne \pm 3$. Reason (R) is true as a general property $\log \left(mn\right)=\log m+\log n$ (for $m,n>0$).

Solution: (a)

Assertion (A) is true because $a^{m+n+m-n}=a^{2m}$. Reason (R) correctly explains this calculation.

Solution: (a)

By Reason (R), $\log x+\log y=\log \left(xy\right)$. Setting this to $0$ means $\log \left(xy\right)=0$, which implies $xy={10}^0=1$. Both A and R are true, and R explains A.

Solution: (a)

Assertion (A) is true because $13=8+4+1=1\cdot 2^3+1\cdot 2^2+0\cdot 2^1+1\cdot 2^0={\left(1101\right)}_2$. Reason (R) is true and accurately explains the binary conversion process.

Solution: (a)

For the outer logarithm to be defined, its argument must be positive, so $\log x>0$. This implies $x>1$. Both A and R are true, and R correctly explains A.