Logarithms

(i) Given,

⇒ 5² = 25

⇒ log₅ (25) = 2.

Hence, logarithmic form is log₅ (25) = 2.

(ii) Given,

⇒ 3^-3 = $\frac{1}{27}$

⇒ log₃ $\Big(\frac{1}{27}\Big)$ = -3.

Hence, logarithmic form is log₃ $\Big(\frac{1}{27}\Big)$ = -3.

(iii) Given,

⇒ $(64)^\frac{1}{3}$ = 4

⇒ log₆₄ 4 = $\frac{1}{3}$.

Hence, logarithmic form is log₆₄ 4 = $\frac{1}{3}$.

(iv) Given,

⇒ 6⁰ = 1

⇒ log₆ 1 = 0.

Hence, logarithmic form is log₆ 1 = 0.

(v) Given,

⇒ 10^-2 = 0.01

⇒ log₁₀ (0.01) = -2.

Hence, logarithmic form is log₁₀ (0.01) = -2.

(vi) Given,

⇒ 4^-1 = $\frac{1}{4}$

⇒ log₄ $\frac{1}{4}$ = -1.

Hence, logarithmic form is log₄ $\frac{1}{4}$ = -1.

(i) Given,

⇒ log₃ 81 = 4

⇒ 3⁴ = 81.

Hence, exponential form is 3⁴ = 81.

(ii) Given,

⇒ log₈ 4 = $\frac{2}{3}$

⇒ $(8)^\frac{2}{3}$ = 4.

Hence, exponential form is $(8)^\frac{2}{3}$ = 4.

(iii) Given,

⇒ log₂ $\frac{1}{8}$ = -3

⇒ 2^-3 = $\frac{1}{8}$.

Hence, exponential form is 2^-3 = $\frac{1}{8}$.

(iv) Given,

⇒ log₁₀ (0.01) = -2

⇒ 10^-2 = 0.01.

Hence, exponential form is 10^-2 = 0.01.

(v) Given,

⇒ log₅ $\Big(\frac{1}{5}\Big)$ = -1

⇒ 5^-1 = $\Big(\frac{1}{5}\Big)$.

Hence, exponential form is 5^-1 = $\Big(\frac{1}{5}\Big)$.

(vi) Given,

⇒ log_a 1 = 0

⇒ a⁰ = 1.

Hence, exponential form is a⁰ = 1.

(i) Let,

⇒ log₂ 64 = x

⇒ 64 = 2^x

⇒ 2⁶ = (2)^x

Equating the exponents,

⇒ x = 6

Hence, log₂ 64 = 6.

(ii) Let,

⇒ log₈ 32 = x

⇒ 32 = 8^x

⇒ 2⁵ = (2³)^x

⇒ 2⁵ = (2)^3x

Equating the exponents,

⇒ 3x = 5

⇒ x = $\frac{5}{3}$.

Hence, log₈ 32 = $\frac{5}{3}$.

(iii) Let,

$$\begin{aligned}\Rightarrow \log_{3} \space {\frac{1}{9}} = x \\[1em] \Rightarrow \frac{1}{9} = 3^x \\[1em] \Rightarrow \frac{1}{3^2} = 3^x \\[1em] \Rightarrow 3^{-2} = 3^x \\[1em]\end{aligned}$$

Equating the exponents,

⇒ x = -2.

Hence, $\log_{3} \space {\frac{1}{9}}$ = -2.

(iv) Let,

⇒ log_0.5 (16) = x

⇒ 16 = 0.5^x

⇒ 2⁴ = $\Big(\frac{1}{2}\Big)^x$

⇒ 2⁴ = (2^-1)^x

⇒ 2⁴ = (2)^-x

Equating the exponents,

⇒ -x = 4

⇒ x = -4.

Hence, log_0.5 (16) = -4.

(v) Let,

⇒ log₂ (0.125) = x

⇒ 0.125 = 2^x

⇒ $\frac{125}{1000}$ = 2^x

⇒ $\frac{1}{8}$ = 2^x

⇒ $\frac{1}{2^3}$ = 2^x

⇒ 2^-3 = 2^x

Equating the exponents,

⇒ x = -3.

Hence, log₂ (0.125) = -3.

(vi) Let,

⇒ log₇ 7 = x

⇒ 7 = 7^x

⇒ 7¹ = 7^x

Equating the exponents,

⇒ x = 1.

Hence, log₇ 7 = 1.

(i) Given,

⇒ log₂ x = -2

⇒ x = 2^-2

⇒ x = $\frac{1}{2^2}$

⇒ x = $\frac{1}{4}$

Hence, x = $\frac{1}{4}$.

(ii) Given,

⇒ log_x 9 = 1

⇒ 9 = x¹

⇒ x = 9.

Hence, x = 9.

(iii) Given,

⇒ log₉ 243 = x

⇒ 243 = 9^x

⇒ 3⁵ = (3²)^x

⇒ 3⁵ = 3^2x

Equating the exponents,

⇒ 2x = 5

⇒ x = $\frac{5}{2}$.

Hence, x = $\frac{5}{2}$.

(iv) Given,

⇒ log₃ x = 0

⇒ x = 3⁰

⇒ x = 1.

Hence, x = 1.

(v) Given,

$$\begin{aligned}\Rightarrow \log_{\sqrt{3}} \space (x − 1) = 2 \\[1em] \Rightarrow (x − 1) = (\sqrt{3})^2 \\[1em] \Rightarrow (x − 1) = 3 \\[1em] \Rightarrow x = 3 + 1 \\[1em] \Rightarrow x = 4\end{aligned}$$

Hence, x = 4.

(vi) Given,

⇒ log₅ (x² − 19) = 3

⇒ (x² − 19) = 5³

⇒ x² − 19 = 125

⇒ x² = 125 + 19

⇒ x² = 144

⇒ x = $\sqrt{144}$

⇒ x = ±12.

Hence, x = ± 12.

(vii) Given,

$$\begin{aligned}\Rightarrow \log_x \space 64 = \frac{3}{2} \\[1em] \Rightarrow 64 = x ^ \frac{3}{2} \\[1em] \Rightarrow 64^{\frac{2}{3}} = x ^{\Big(\frac{3}{2} \times \frac{2}{3} \Big)} \\[1em] \Rightarrow x = 64^{\frac{2}{3}} \\[1em] \Rightarrow x = (4^3)^\frac{2}{3} \\[1em] \Rightarrow x = 4^2 \\[1em] \Rightarrow x = 16.\end{aligned}$$

Hence, x = 16.

(viii) Given,

⇒ log₂ (x² − 9) = 4

⇒ (x² − 9) = 2⁴

⇒ x² − 9 = 16

⇒ x² = 16 + 9

⇒ x² = 25

⇒ x = $\sqrt{25}$

⇒ x = ±5

Hence, x = ±5.

(ix) Given,

⇒ log_x (0.008) = −3

⇒ 0.008 = x⁻³

⇒ $\frac{8}{1000}$ = x⁻³

⇒ $\frac{1}{125}$ = x⁻³

⇒ $\frac{1}{5^3}$ = x⁻³

⇒ 5⁻³ = x⁻³

Equating the bases,

⇒ x = 5.

Hence, x = 5.

Given,

⇒ log₁₀ x = p and log₁₀ y = q

⇒ x = 10^p and y = 10^q

⇒ x × y = 10^p × 10^q

⇒ xy = (10)^{p + q}.

Hence, proved that xy = (10)^{p + q}.

Given,

⇒ log₁₀ x = a and log₁₀ y = b

⇒ x = 10^a and y = 10^b

(i) Given,

⇒ 10^{a + 1}

⇒ 10^a × 10¹

⇒ x × 10

⇒ 10x.

Hence, 10^{a + 1} = 10x.

(ii) Given,

⇒ 10^2b

⇒ (10^b)²

⇒ y².

Hence, 10^2b = y².

(iii) Given,

⇒ log₁₀ P = 2a − b

⇒ P = 10^{(2a − b)}

⇒ P = 10^2a × 10^−b

⇒ P = (10^a)² × (10^b)^-1

⇒ P = (x)² × (y)^-1

⇒ P = $\frac{x^2}{y}$.

Hence, P = $\frac{x^2}{y}$.

Given,

⇒ 2 log 5 + log 8 − $\frac{1}{2}$ log 4

$$\begin{aligned}\Rightarrow 2 \log \space 5 + \log \space 8 − \frac{1}{2} \text{ log} 4 \\[1em] \Rightarrow \log \space (5)^2 + \log \space 8 − \log \space(4)^{\frac{1}{2}} \\[1em] \Rightarrow \log \space 25 + \log \space 8 − \log \space \sqrt{4} \\[1em] \Rightarrow \log \space 25 + \log \space 8 − \log \space 2 \\[1em] \Rightarrow \log \space (25 × 8) − \log \space 2 \\[1em] \Rightarrow \log \space (200) − \log \space 2 \\[1em] \Rightarrow \log \space \Big(\frac{200}{2}\Big) \\[1em] \Rightarrow \log \space 100 \\[1em] \Rightarrow \log \space 10^2 \\[1em] \Rightarrow 2\log \space 10 \\[1em] \Rightarrow 2 \times 1 \\[1em] \Rightarrow 2.\end{aligned}$$

Hence, 2 log 5 + log 8 − $\frac{1}{2}$ log 4 = 2.

Given,

⇒ log 8 + log 25 + 2 log 3 − log 18

⇒ log (8 × 25) + log 3² - log 18

⇒ log (200) + log 9 - log 18

⇒ log (200 × 9) - log 18

⇒ log (1800) - log 18

⇒ log $\Big(\frac{1800}{18}\Big)$

⇒ log 100

⇒ log 10²

⇒ 2 log 10

⇒ 2 × 1

⇒ 2.

Hence, log 8 + log 25 + 2 log 3 − log 18 = 2.

Given,

⇒ 5 log 2 + $\frac{3}{2}$ log 25 + $\frac{1}{2}$ log 49 − log 28

$$\begin{aligned}\Rightarrow 5 \log \space 2 + \frac{3}{2} \log \space 25 + \frac{1}{2} \log \space 49 − \log \space 28 \\[1em] \Rightarrow \log \space 2^5 + \log \space 25^\frac{3}{2} + \log \space 49^\frac{1}{2} − \log \space 28 \\[1em] \Rightarrow \log \space 32 + \log \space (5^2)^{\frac{3}{2}} + \log \space \sqrt{49} − \log \space 28 \\[1em] \Rightarrow \log \space 32 + \log \space 5^3 + \log \space 7 − \log \space 28 \\[1em] \Rightarrow \log \space 32 + \log \space 125 + \log \space 7 − \log \space 28 \\[1em] \Rightarrow \log \space (32 × 125 × 7) − \log \space 28 \\[1em] \Rightarrow \log \space 28000 − \log \space 28 \\[1em] \Rightarrow \log \space \Big(\frac{28000}{28}\Big) \\[1em] \Rightarrow \log \space 1000 \\[1em] \Rightarrow \log \space 10^3 \\[1em] \Rightarrow 3 \log \space 10 \\[1em] \Rightarrow 3 \times 1 \\[1em] \Rightarrow 3.\end{aligned}$$

Hence, 5 log 2 + $\frac{3}{2}$ log 25 + $\frac{1}{2}$ log 49 − log 28 = 3.

Given,

$$\begin{aligned}2 \log 2 + \log 5 − \frac{1}{2} \log 36 − \log \frac{1}{30}\\[1em] \Rightarrow \log \space 2^2 + \log \space 5 − \log \space 36^{\frac{1}{2}} − \log \space \frac{1}{30} \\[1em] \Rightarrow \log \space 4 + \log \space 5 − \log \space \sqrt{36} − (\log \space 1 - \log \space 30) \\[1em] \Rightarrow \log \space (4 \times 5) − \log \space 6 - (0 - \log \space 30) \\[1em] \Rightarrow \log \space 20 + \log \space 30 − \log \space 6 \\[1em] \Rightarrow \log \space (20 \times 30) − \log \space 6 \\[1em] \Rightarrow \log \space \Big(\frac{600}{6}\Big) \\[1em] \Rightarrow \log \space 100 \\[1em] \Rightarrow \log \space 10^2 \\[1em] \Rightarrow 2\log \space 10 \\[1em] \Rightarrow 2 \times 1 \\[1em] \Rightarrow 2.\end{aligned}$$

Hence, 2 log 2 + log 5 − $\frac{1}{2}$ log 36 − log $\frac{1}{30}$ = 2.

Given,

⇒ log (1.2) + 2 log (0.75) − log (6.75)

⇒ log (1.2) + log (0.75)² − log (6.75)

⇒ log (1.2) + log (0.5625) − log (6.75)

⇒ log (1.2 × 0.5625) − log (6.75)

⇒ log (0.675) − log (6.75)

⇒ log $\Big(\frac{0.675}{6.75}\Big)$

⇒ log (0.1)

⇒ log $\Big(\frac{1}{10}\Big)$

⇒ log₁₀ 10^-1

⇒ -1 log₁₀ 10

⇒ -1 × 1

⇒ -1.

Hence, log (1.2) + 2 log (0.75) − log (6.75) = -1.

Given,

$$\begin{aligned}\Rightarrow \log \space 5 + 16 \log \space \Big(\frac{625}{6}\Big) + 12 \log \space \Big(\frac{4}{375}\Big) + 7 \log \space \Big(\frac{81}{1250}\Big) \\[1em] \Rightarrow \log \space 5 + \log \space \Big(\frac{625}{6}\Big)^{16} + \log \space \Big(\frac{4}{375}\Big)^{12} + \log \space \Big(\frac{81}{1250}\Big)^{7} \\[1em] \Rightarrow \log \space 5 + \log \space \Big(\frac{5^4}{2 \times 3}\Big)^{16} + \log \space \Big(\frac{2^2}{3 \times 5^3 }\Big)^{12} + \log \space \Big(\frac{3^4}{2 \times 5^4}\Big)^{7} \\[1em] \Rightarrow \log \space 5 + \log \space \Big(\frac{5^{(4 \times 16)}}{2^{16} \times 3^{16}}\Big) + \log \space \Big(\frac{2^{2 \times 12}}{3^{12} \times 5^{3 \times 12}}\Big) + \log \space \Big(\frac{3^{4 \times 7}}{2^7 \times 5^{4 \times 7}}\Big) \\[1em] \Rightarrow \log \space 5 + \log \space \Big(\frac{5^{64}}{2^{16} \times 3^{16}}\Big) + \log \space \Big(\frac{2^{24}}{3^{12} \times 5^{36} }\Big) + \log \space \Big(\frac{3^{28}}{2^{7} \times 5^{28}}\Big) \\[1em] \Rightarrow \log \space \Big[\frac{5 \times 5^{64} \times 2^{24} \times 3^{28} }{2^{16} \times 3^{16} \times 3^{12} \times 5^{36} \times 2^{7} \times 5^{28}}\Big] \\[1em] \Rightarrow \log \space \Big[\frac{5^{64 + 1} \times 2^{24} \times 3^{28} }{2^{16 + 7} \times 3^{16 + 12} \times 5^{36 + 28}}\Big] \\[1em] \Rightarrow \log \space \Big[\frac{5^{65} \times 2^{24} \times 3^{28} }{2^{23} \times 3^{28} \times 5^{64}}\Big] \\[1em] \Rightarrow \log \space [5^{65 - 64} \times 2^{24 - 23} \times 3^{28 - 28} ] \\[1em] \Rightarrow \log \space [5^1 \times 2^1 \times 3^0] \\[1em] \Rightarrow \log \space [5 \times 2 \times 1] \\[1em] \Rightarrow \log \space 10 \\[1em] \Rightarrow 1.\end{aligned}$$

Hence, $\log \space \Big(\frac{81}{8}\Big) + 2 \log \space \frac{2}{3} − 3 \log \space \frac{3}{2} + \log \space \frac{3}{4}$ = 1.

Given,

$$\begin{aligned}\Rightarrow 3 \log \space 2 − \frac{1}{3} \log \space 27 + \log \space 12 − \log \space 4 + 3 \log \space 5 \\[1em] \Rightarrow \log \space 2^3 − \log \space 27^{\frac{1}{3}} + \log \space 12 − \log \space 4 + \log \space 5 ^{3} \\[1em] \Rightarrow \log \space 8 − \log \space \sqrt[3]{27} + \log \space 12 − \log \space 4 + \log \space 125 \\[1em] \Rightarrow \log \space 8 − \log \space 3 + \log \space 12 − \log \space 4 + \log \space 125 \\[1em] \Rightarrow \log \space 8 + \log \space 12 + \log \space 125 − \log \space 4 − \log \space 3 \\[1em] \Rightarrow (\log \space 8 + \log \space 12 + \log \space 125) − (\log \space 4 + \log \space 3) \\[1em] \Rightarrow \log \space (8 \times 12 \times 125) - \log \space (4 \times 3) \\[1em] \Rightarrow \log \space (12000) - \log \space (12) \\[1em] \Rightarrow \log \space \Big(\frac{12000}{12}\Big) \\[1em] \Rightarrow \log \space 1000 \\[1em] \Rightarrow \log \space 10^3 \\[1em] \Rightarrow 3\log \space 10 \\[1em] \Rightarrow 3 \times 1 \\[1em] \Rightarrow 3.\end{aligned}$$

Hence, $3 \log \space 2 − \frac{1}{3} \log \space 27 + \log \space 12 − \log \space 4 + 3 \log \space 5$ = 3.

Given,

$\Rightarrow \log \space \Big(\frac{81}{8}\Big) + 2 \log \space \Big(\frac{2}{3}\Big) − 3 \log \space \Big(\frac{3}{2}\Big) + \log \space \Big(\frac{3}{4}\Big)$

⇒ log 81 - log 8 + 2(log 2 - log 3) - 3 (log 3 - log2) + (log 3 - log 4)

⇒ log 81 - log 8 + 2log 2 - 2log 3 - 3log 3 + 3log 2 + log 3 - log 4

⇒ log 81 - log 8 + log 2² - log 3² - log 3³ + log 2³ + log 3 - log 4

⇒ log 81 - log 8 + log 4 - log 9 - log 27 + log 8 + log 3 - log 4

⇒ log 81 + log 4 + log 8 + log 3 - log 4 - log 8 - log 9 - log 27

⇒ (log 81 + log 4 + log 8 + log 3) - (log 4 + log 8 + log 9 + log 27)

⇒ log (81 × 4 × 8 × 3) - log (8 × 9 × 27 × 4)

⇒ log (7776) - log (7776)

⇒ 0.

Hence, $\log \space \Big(\frac{81}{8}\Big) + 2 \log \space \Big(\frac{2}{3}\Big) − 3 \log \space \Big(\frac{3}{2}\Big) + \log \space \Big(\frac{3}{4}\Big)$ = 0.

Given,

$\log_{10} \space \Big(\frac{x^2y^3}{z}\Big)$

⇒ log₁₀ x² y³ - log₁₀ z

⇒ log₁₀ x² + log₁₀ y³ - log₁₀ z

⇒ 2 log₁₀ x + 3 log₁₀ y - log₁₀ z.

Hence, log₁₀ $\Big(\frac{x^2y^3}{z}\Big)$ = 2 log₁₀ x + 3 log₁₀ y - log₁₀ z.

Given,

$$\begin{aligned}\Rightarrow \log_{10} \space \Big(\frac{\sqrt{pq^3}}{r^2s}\Big) \\[1em] \Rightarrow \log_{10} \space \sqrt{pq^3} - \log_{10} \space {r^2s} \\[1em] \Rightarrow \log_{10} \space ({pq^3})^{\frac{1}{2}} - (\log_{10} \space {r^2} + \log_{10} \space {s}) \\[1em] \Rightarrow \frac{1}{2}\log_{10} \space {pq^3} - (2\log_{10} \space {r} + \log_{10} \space {s}) \\[1em] \Rightarrow \frac{1}{2}(\log_{10} \space {p} + \log_{10} \space {q^3}) - 2\log_{10} \space {r} - \log_{10} \space {s} \\[1em] \Rightarrow \frac{1}{2}(\log_{10} \space {p} + 3\log_{10} \space {q}) - 2\log_{10} \space {r} - \log_{10} \space {s} \\[1em]\end{aligned}$$

Hence, $\log_{10} \space \Big(\frac{\sqrt{pq^3}}{r^2s}\Big) = \frac{1}{2}(\log_{10} \space {p} + 3\log_{10} \space {q}) - 2\log_{10} \space {r} - \log_{10} \space {s}$.

Given,

⇒ 2 log₁₀ 8 + log₁₀ 36 − log₁₀ (1.5) − 3 log₁₀ 2

⇒ log₁₀ 8² + log₁₀ 36 − log₁₀ (1.5) − log₁₀ 2³

⇒ log₁₀ 64 + log₁₀ 36 − log₁₀ (1.5) − log₁₀ 8

⇒ log₁₀ 64 + log₁₀ 36 − (log₁₀ (1.5) + log₁₀ 8)

⇒ log₁₀ (64 × 36) - log₁₀(1.5 × 8)

⇒ log₁₀ (2304) - log₁₀(12)

⇒ $\log_{10} \space {\Big(\frac{2304}{12}\Big)}$

⇒ log₁₀ 192

Hence, 2 log₁₀ 8 + log₁₀ 36 − log₁₀ (1.5) − 3 log₁₀ 2 = log₁₀ 192.

Given,

⇒ 2 log₁₀ 5 + 2 log₁₀ 3 − log₁₀ 2 + 1

⇒ log₁₀ 5² + log₁₀ 3² − log₁₀ 2 + 1

⇒ log₁₀ 25 + log₁₀ 9 − log₁₀ 2 + log₁₀ 10

⇒ log₁₀ 25 + log₁₀ 9 + log₁₀ 10 - log₁₀ 2

⇒ log₁₀ (25 × 9 × 10) − log₁₀ 2

⇒ log₁₀ (2250) − log₁₀ 2

⇒ $\log_{10} \space {\Big(\frac{2250}{2}\Big)}$

⇒ log₁₀ 1125.

Hence, 2 log₁₀ 5 + 2 log₁₀ 3 − log₁₀ 2 + 1 = log₁₀ 1125.

Given,

$$\begin{aligned}\Rightarrow 2 + \frac{1}{2} \log_{10} \space 9 − 2 \log_{10} \space 5 \\[1em] \Rightarrow 2\log_{10} \space 10 + \log_{10} \space 9^{\frac{1}{2}} − \log_{10} \space 5^2 \\[1em] \Rightarrow \log_{10} \space 10^2 + \log_{10} \space (3^2)^{\frac{1}{2}} − \log_{10} \space 25 \\[1em] \Rightarrow \log_{10} \space100 + \log_{10} \space 3 − \log_{10} \space 25 \\[1em] \Rightarrow \log_{10} \space {{(100 \times 3)}} − \log_{10} \space 25 \\[1em] \Rightarrow \log_{10} \space {300} − \log_{10} \space 25 \\[1em] \Rightarrow \log_{10} \space {\Big(\frac{300}{25}\Big)} \\[1em] \Rightarrow \log_{10} \space 12.\end{aligned}$$

Hence, $2 + \frac{1}{2} \log_{10} \space 9 − 2 \log_{10} \space 5$ = log₁₀ 12.

Given,

$$\begin{aligned}\Rightarrow \frac{1}{2} \log_{10} \space 9 + \frac{1}{4} \log_{10} \space 81 + 2 \log_{10} \space 6 − \log_{10} \space 12 \\[1em] \Rightarrow \log_{10} 9^{\frac{1}{2}} + \log_{10} \space 81^{\frac{1}{4}} + \log_{10} \space6^2 − \log_{10} \space 12 \\[1em] \Rightarrow \log_{10} \sqrt{9} + \log_{10} \space (3^4)^{\frac{1}{4}} + \log_{10} \space 36 − \log_{10} \space 12 \\[1em] \Rightarrow \log_{10} \space 3 + \log_{10} \space 3 + \log_{10} \space 36 − \log_{10} \space 12 \\[1em] \Rightarrow \log_{10} \space {(3 \times 3 \times 36)} − \log_{10} \space 12 \\[1em] \Rightarrow \log_{10} 324 − \log_{10} 12 \\[1em] \Rightarrow \log_{10} \frac{324}{12} \\[1em] \Rightarrow \log_{10} 27.\end{aligned}$$

Hence, $\frac{1}{2} \log_{10} 9 + \frac{1}{4} \log_{10} 81 + 2 \log_{10} 6 − \log_{10} 12$ = log₁₀ 27.

Given,

$\Rightarrow 2 \log_{10} \space \Big(\frac{11}{13}\Big) + \log_{10} \space \Big(\frac{130}{77}\Big) − \log_{10} \space \Big(\frac{55}{91}\Big)$

⇒ 2(log₁₀ 11 - log₁₀ 13) + (log₁₀ 130 - log₁₀77) - (log₁₀ 55 - log₁₀ 91)

⇒ 2log₁₀ 11 - 2log₁₀ 13 + log₁₀ 130 - log₁₀77 - log₁₀ 55 + log₁₀ 91

⇒ log₁₀ 11² - log₁₀ 13² + log₁₀ 130 - log₁₀77 - log₁₀ 55 + log₁₀ 91

⇒ log₁₀ 121 - log₁₀ 169 + log₁₀ 130 - log₁₀77 - log₁₀ 55 + log₁₀ 91

⇒ log₁₀ 121 + log₁₀ 130 + log₁₀ 91 - log₁₀ 169 - log₁₀77 - log₁₀ 55

⇒ log₁₀ (121 × 130 × 91) - log₁₀ (169 × 77 × 55)

⇒ $\log_{10} \space {\frac{121 \times 130 \times 91}{169 \times 77 \times 55}}$

⇒ $\log_{10} \space {\frac{26}{13}}$

⇒ log₁₀ 2.

Hence, $2 \log_{10} \space \Big(\frac{11}{13}\Big) + \log_{10} \space \Big(\frac{130}{77}\Big) − \log_{10} \space \Big(\frac{55}{91}\Big)$ = log₁₀ 2.

Given,

$$\begin{aligned}\Rightarrow 1 - \frac{1}{3} \log_{10} 64 \\[1em] \Rightarrow \log_{10} 10 − \log_{10} 64^{\frac{1}{3}} \\[1em] \Rightarrow \log_{10} 10 − \log_{10} \sqrt[3]{64} \\[1em] \Rightarrow \log_{10} 10 − \log_{10} 4 \\[1em] \Rightarrow \log_{10} \frac{10}{4} \\[1em] \Rightarrow \log_{10} \frac{5}{2} \\[1em] \Rightarrow \log_{10} \space 2.5\end{aligned}$$

Hence, $1 − \frac{1}{3} \log_{10} 64$ = log₁₀ 2.5

Given,

$$\begin{aligned}\Rightarrow \frac{\log \space 81}{\log \space 27} \\[1em] \Rightarrow \frac{\log \space 3^4}{\log \space 3^3} \\[1em] \Rightarrow \frac{4\log \space 3}{3\log \space 3} \\[1em] \Rightarrow \frac{4}{3}.\end{aligned}$$

Hence, $\frac{\log \space 81}{\log \space 27} = \frac{4}{3}$.

Given,

$$\begin{aligned}\Rightarrow \frac{\log \space 128}{\log \space 32} \\[1em] \Rightarrow \frac{\log \space 2^7}{\log \space 2^5} \\[1em] \Rightarrow \frac{7\log \space 2}{5\log \space 2} \\[1em] \Rightarrow \frac{7}{5}.\end{aligned}$$

Hence, $\frac{\log \space 128}{\log \space 32} = \frac{7}{5}$.

Given,

$$\begin{aligned}\Rightarrow \frac{\log \space 27}{\log \space \sqrt{3}} \\[1em] \Rightarrow \frac{\log \space 3^3}{\log \space {3}^{\frac{1}{2}}} \\[1em] \Rightarrow \frac{3\log \space 3}{{\frac{1}{2}}\log \space 3} \\[1em] \Rightarrow \frac{3}{{\frac{1}{2}}} \\[1em] \Rightarrow 3 \times 2 \\[1em] \Rightarrow 6.\end{aligned}$$

Hence, $\frac{\log \space 27}{\log \space \sqrt{3}}$ = 6.

Given,

$$\begin{aligned}\Rightarrow \frac{\log \space 9 - \log \space 3}{\log \space 27} \\[1em] \Rightarrow \frac{\log \space {\frac{9}{3}}}{\log \space 3^3} \\[1em] \Rightarrow \frac{\log \space 3}{3\log \space 3} \\[1em] \Rightarrow \frac{1}{3}.\end{aligned}$$

Hence, $\frac{\log \space 9 - \log \space 3}{\log \space 27} = \frac{1}{3}$.

(i) Given,

⇒ log 12

⇒ log (3 × 2 × 2)

⇒ log 3 + log 2 + log 2

⇒ 0.4771 + 0.3010 + 0.3010

⇒ 0.4771 + 0.3010 + 0.3010

⇒ 1.0791

Hence, log 12 = 1.0791.

(ii) Given,

⇒ log 25

⇒ log 5²

⇒ 2 log 5

⇒ $2\text{ log }\frac{10}{2}$

⇒ 2(log 10 - log 2)

⇒ 2(1 - 0.3010)

⇒ 2(0.699)

⇒ 1.398

Hence, log 25 = 1.398

(iii) Given,

$$\begin{aligned}\Rightarrow \log \space \sqrt{18} \\[1em] \Rightarrow \log \space {18}^{\frac{1}{2}} \\[1em] \Rightarrow \frac{1}{2}\log \space 18 \\[1em] \Rightarrow \frac{1}{2}\log \space (3 \times 3 \times 2) \\[1em] \Rightarrow \frac{1}{2}(\log \space 3 + \log \space 3 + \log \space 2) \\[1em] \Rightarrow \frac{1}{2}(0.4771 + 0.4771 + 0.3010) \\[1em] \Rightarrow \frac{1}{2}(1.2552) \\[1em] \Rightarrow 0.6276\end{aligned}$$

Hence, $\log \sqrt{18}$ =0.6276.

(iv) Given,

$\Rightarrow \log \space \Big(\frac{9}{4}\Big)$

⇒ log 9 - log 4

⇒ log 3² - log 2²

⇒ 2 log 3 - 2 log 2

⇒ 2 × 0.4771 - 2 × 0.3010

⇒ 0.9542 - 0.6020

⇒ 0.3522

Hence, $\log \Big(\frac{9}{4}\Big)$ value is 0.3522.

Given,

$\Rightarrow \Big(\log \space \frac{75}{16} - 2 \log \space \frac{5}{9} + \log \space \frac{32}{243}\Big)$

⇒ log 75 - log 16 - 2 (log 5 - log 9) + log 32 - log 243

⇒ log (25 × 3) - log 16 - 2 (log 5 - log 9) + log 2⁵ - log 3⁵

⇒ log 25 + log 3 - log 2⁴ - 2log 5 + 2log 9 + 5log 2 - 5log 3

⇒ log 25 + log 3 - 4log 2 - log 5² + 2log 3² + 5log 2 - 5log 3

⇒ log 25 + log 3 - 4log 2 - log 25 + 4log 3 + 5log 2 - 5log 3

⇒ log 25 - log 25 - 4log 2 + 5log 2 + log 3 + 4log 3 - 5log 3

⇒ log 2 + log 3 + 5log 3 - 5log 3

⇒ log 2

⇒ 0.3010

Hence, $\Big(\log \space \frac{75}{16} - 2 \log \space \frac{5}{9} + \log \space \frac{32}{243}\Big)$ = 0.3010.

Given,

⇒ log 8 = 0.9030

⇒ log 2³ = 0.9030

⇒ 3log 2 = 0.9030

⇒ log 2 = $\frac{0.9030}{3}$

⇒ log 2 = 0.3010

(i) Given,

⇒ log 4

⇒ log 2²

⇒ 2log 2

⇒ 2 × (0.3010)

⇒ 0.6020

Hence, log 4 = 0.6020.

(ii) Given,

$$\begin{aligned}\Rightarrow \log \space \sqrt{32} \\[1em] \Rightarrow \log \space {32}^{\frac{1}{2}} \\[1em] \Rightarrow \frac{1}{2}\log \space {32} \\[1em] \Rightarrow \frac{1}{2}\log \space {(8 \times 4)} \\[1em] \Rightarrow \frac{1}{2}(\log \space {8} + \log \space {4}) \\[1em] \Rightarrow \frac{1}{2}(0.9030 + \log \space {2^2}) \\[1em] \Rightarrow \frac{1}{2}(0.9030 + 2\log \space {2}) \\[1em] \Rightarrow \frac{1}{2}[0.9030 + 2 \times (0.3010)] \\[1em] \Rightarrow \frac{1}{2}(0.9030 + 0.6020) \\[1em] \Rightarrow \frac{1}{2} \times 1.5050 \\[1em] \Rightarrow 0.7525\end{aligned}$$

Hence, $\log \sqrt{32}$ = 0.7525.

(iii) Given,

⇒ log (0.125)

⇒ $\log \space {\Big(\frac{125}{1000}\Big)}$

⇒ $\log \space {\Big(\frac{1}{8}\Big)}$

⇒ log 1 - log 8

⇒ 0 - 0.9030

⇒ -0.9030

Hence, log (0.125) = -0.9030.

Given,

⇒ log 27 = 1.4313

⇒ log 3³ = 1.4313

⇒ 3log 3 = 1.4313

⇒ log 3 = $\frac{1.4313}{3}$

⇒ log 3 = 0.4771

(i) Given,

⇒ log 9

⇒ log 3²

⇒ 2log 3

⇒ 2 × (0.4771)

⇒ 0.9542

Hence, log 9 = 0.9542.

(ii) Given,

⇒ log 30

⇒ log (3 × 10)

⇒ log 3 + log 10

⇒ log 3 + 1

⇒ 0.4771 + 1

⇒ 1.4771.

Hence, log 30 = 1.4771.

Given,

⇒ log (1 + 2 + 3) = log 1 + log 2 + log 3

Solving the R.H.S,

⇒ log 1 + log 2 + log 3

⇒ log (1 × 2 × 3)

⇒ log 6

⇒ log (1 + 2 + 3)

Since,

L.H.S = R.H.S.

Hence, proved that log (1 + 2 + 3) = log 1 + log 2 + log 3.

Given,

⇒ log (m + n) = log m + log n

⇒ log (m + n) = log (mn)

⇒ (m + n) = (mn)

⇒ m - mn + n = 0

⇒ m(1 - n) + n = 0

⇒ m(1 - n) = - n

⇒ m = $\frac{-n}{(1 - n)}$

⇒ m = $\frac{-n}{-(n - 1)}$

⇒ m = $\frac{n}{(n - 1)}$.

Hence, proved that $m = \frac{n}{n - 1}$.

Given,

$$\begin{aligned}\Rightarrow \log \space \Big(\frac{a + b}{2}\Big) = \frac{1}{2} (\log \space a + \log \space b) \\[1em] \Rightarrow \log \space \Big(\frac{a + b}{2}\Big) = \frac{1}{2} \log \space ab \\[1em] \Rightarrow \log \space \Big(\frac{a + b}{2}\Big) = \log \space {ab} ^{\frac{1}{2}} \\[1em] \Rightarrow \Big(\frac{a + b}{2}\Big) = {ab} ^{\frac{1}{2}} \\[1em] \Rightarrow \frac{1}{2} (a + b) = \sqrt{ab}.\end{aligned}$$

Hence, proved that $\frac{1}{2} (a + b) = \sqrt{ab}$.

Given,

⇒ log (x + 2) + log (x − 2) = log 5

⇒ log [(x + 2) × (x − 2)] = log 5

⇒ log [(x² - 2²)] = log 5

⇒ log [(x² - 4)] = log 5

⇒ (x² - 4) = 5

⇒ x² = 5 + 4

⇒ x² = 9

⇒ x = $\sqrt{9}$

⇒ x = ± 3.     

x cannot be equal to -3 as that will make x + 2 = -3 + 2 = -1, that is negative, and argument of a logarithm cannot be negative.

Hence, the value of x = 3.

Given,

$$\begin{aligned}\Rightarrow \log \space (x + 4) - \log \space (x - 4) = \log \space 2 \\[1em] \Rightarrow \log \space \frac{(x + 4)}{(x - 4)} = \log \space 2 \\[1em] \Rightarrow \frac{(x + 4)}{(x - 4)} = 2 \\[1em] \Rightarrow (x + 4) = (x - 4) \times 2 \\[1em] \Rightarrow x + 4 = 2x - 8 \\[1em] \Rightarrow 2x - x = 8 + 4 \\[1em] \Rightarrow x = 12 .\end{aligned}$$

Hence, the value of x = 12.

Given,

$$\begin{aligned}\Rightarrow \log \space (x + 3) − \log \space (x − 3) = 1 \\[1em] \Rightarrow \log \space \frac{(x + 3)} {(x − 3)} = \log \space 10 \\[1em] \Rightarrow \frac{(x + 3)} {(x − 3)} = 10 \\[1em] \Rightarrow (x + 3) = (x − 3) \times 10 \\[1em] \Rightarrow x + 3 = 10x − 30 \\[1em] \Rightarrow 10x - x = 3 + 30 \\[1em] \Rightarrow 9x = 33 \\[1em] \Rightarrow x = \frac{33}{9} \\[1em] \Rightarrow x = \frac{11}{3}.\end{aligned}$$

Hence, the value of x = $\frac{11}{3}$.

Given,

⇒ log (x² − 21) = 2

⇒ log (x² − 21) = 2 log 10

⇒ log (x² − 21) = log 100

⇒ (x² − 21) = 100

⇒ x² = 100 + 21

⇒ x² = 121

⇒ x = $\sqrt{121}$

⇒ x = ±11.

Hence, the value of x = ±11.

Given,

⇒ 2 log x + 1 = log 250

⇒ log x² + log 10 = log 250

⇒ log (10 × x²) = log 250

⇒ (10 × x²) = 250

⇒ x² = $\frac{250}{10}$

⇒ x² = 25

⇒ x = $\sqrt{25}$

⇒ x = ±5.

x cannot be equal to -5 as the argument of a logarithm cannot be negative.

Hence, the value of x = 5.

Given,

$$\begin{aligned}\Rightarrow \frac{\log \space x}{\log \space 5} = \frac{\log \space 9}{\log \space \Big(\frac{1}{3}\Big)} \\[1em] \Rightarrow \frac{\log \space x}{\log \space 5} = \frac{\log \space 3^2}{\log \space 3 ^{-1}} \\[1em] \Rightarrow \frac{\log \space x}{\log \space 5} = \frac{2\log \space 3}{-1\log \space 3} \\[1em] \Rightarrow \frac{\log \space x}{\log \space 5} = -2 \\[1em] \Rightarrow \log \space x = -2 \times \log \space 5 \\[1em] \Rightarrow \log \space x = \log \space 5 ^{-2} \\[1em] \Rightarrow \log \space x = \log \space \Big(\frac{1}{25}\Big) \\[1em] \Rightarrow x = \frac{1}{25}.\end{aligned}$$

Hence, the value of x = $\frac{1}{25}$.

Given,

$\Rightarrow \log \space 7 - \log \space 2 + \log \space 16 - 2 \log \space 3 - \log \space \frac{7}{45} = 1 + \log \space n$

⇒ log 7 - log 2 + log 2⁴ - 2log 3 - (log 7 - log 45) = log 10 + log n

⇒ log 7 - log 2 + 4log 2 - 2log 3 - log 7 + log 45 = log 10n

⇒ log 7 - log 7- log 2 + 4log 2 - 2log 3 + log 45 = log 10n

⇒ 3log 2 - 2log 3 + log (9 × 5) = log 10n

⇒ 3log 2 - 2log 3 + log 9 + log 5 = log 10n

⇒ log 2³ - 2log 3 + log 3² + log 5 = log 10n

⇒ log 8 - 2log 3 + 2log 3 + log 5 = log 10n

⇒ log 8 + log 5 = log 10n

⇒ log (8 × 5) = log 10n

⇒ log 40 = log 10n

⇒ 10n = 40

⇒ n = $\frac{40}{10}$

⇒ n = 4.

Hence, the value of n = 4.

Given,

$$\begin{aligned}\Rightarrow R = \sqrt {\frac{3V}{\pi h}} \\[1em] \text{Taking log on Both sides,} \\[1em] \Rightarrow \log \space R = \log \space {\sqrt{\frac{3V}{\pi h}}} \\[1em] \Rightarrow \log \space R = \log \space {\Big(\frac{3V}{\pi h}\Big)}^{\frac{1}{2}} \\[1em] \Rightarrow \log \space R = \frac{1}{2} \log \space {\Big(\frac{3V}{\pi h}\Big)} \\[1em] \Rightarrow \log \space R = \frac{1}{2} (\log \space {3V} - \log \space {\pi h}) \\[1em] \Rightarrow \log \space R = \frac{1}{2} [\log \space {3} + \log \space {V} - (\log \space {\pi} + \log \space { h})] \\[1em] \Rightarrow \log \space R = \frac{1}{2} (\log \space {3} + \log \space {V} - \log \space {\pi} - \log \space { h})\end{aligned}$$

Hence, logarithmic equation is $\log \space R = \frac{1}{2} (\log \space {3} + \log \space {V} - \log \space {\pi} - \log \space { h})$.

Given,

$$\begin{aligned}\Rightarrow x = ab \sqrt{\frac{a - b}{a + b}} \\[1em] \text{Taking log on Both sides,} \\[1em] \Rightarrow \log \space x = \log \space \Big(ab \sqrt{\frac{a - b}{a + b}}\Big) \\[1em] \Rightarrow \log \space x = \log \space ab + \log \space\sqrt{\frac{a - b}{a + b}} \\[1em] \Rightarrow \log \space x = \log \space a + \log \space b + \log \space \Big({\frac{a - b}{a + b}}\Big)^{\frac{1}{2}} \\[1em] \Rightarrow \log \space x = \log \space a + \log \space b + \frac{1}{2} \log \space {\frac{a - b}{a + b}} \\[1em] \Rightarrow \log \space x = \log \space a + \log \space b + \frac{1}{2} [\log \space ({a - b}) - \log \space ({a + b})] \\[1em]\end{aligned}$$

Hence, logarithmic equation is

$\log \space x = \log \space a + \log \space b + \frac{1}{2} [\log \space ({a - b}) - \log \space ({a + b})]$

Given,

$$\begin{aligned}\Rightarrow \sqrt[3]{64} = 4 \\[1em] \Rightarrow (64)^{\frac{1}{3}} = 4 \\[1em] \Rightarrow \log_{64} \space 4 = \frac{1}{3}.\end{aligned}$$

Hence, option 3 is the correct option.

Given,

⇒ log₃ 243 = 5

⇒ 243 = 3⁵

⇒ 3⁵ = 243.

Hence, option 2 is the correct option.

Given,

$\Rightarrow \log_{\sqrt{2}} \space \Big(4\sqrt{2}\Big)$

Let,

$$\begin{aligned}\Rightarrow \log_{\sqrt{2}} \Big(4\sqrt{2}\Big) = y \\[1em] \Rightarrow (\sqrt{2})^{y} = 4\sqrt{2} \\[1em] \Rightarrow [(2)^{\frac{1}{2}}]^{y} = 2^2 \times 2^{\frac{1}{2}} \\[1em] \Rightarrow (2)^{\frac{y}{2}} = 2^{2 + \frac{1}{2}} \\[1em] \Rightarrow (2)^{\frac{y}{2}} = 2^{\frac{4 + 1}{2}} \\[1em] \Rightarrow (2)^{\frac{y}{2}} = 2^{\frac{5}{2}} \\[1em] \Rightarrow \frac{y}{2} = \frac{5}{2} \\[1em] \Rightarrow y = \frac{5}{2} \times 2 \\[1em] \Rightarrow y = 5.\end{aligned}$$

Hence, option 4 is the correct option.

Given,

$\Rightarrow \log_3 \space \Big(\frac{1}{27}\Big)$

Let,

$$\begin{aligned}\Rightarrow \log_3 \space \Big(\frac{1}{27}\Big) = x \\[1em] \Rightarrow \Big(\frac{1}{27}\Big) = 3^x \\[1em] \Rightarrow 3^{-3} = 3^x \\[1em] \Rightarrow x = -3.\end{aligned}$$

Hence, option 4 is the correct option.

Given,

⇒ log 5 + 2log 3

⇒ log 5 + log 3²

⇒ log 5 + log 9

⇒ log (5 × 9)

⇒ log 45.

Hence, option 2 is the correct option.

Given,

⇒ log (1 × 2 × 3)

⇒ log 1 + log 2 + log 3.

Hence, option 3 is the correct option.

Let,

⇒ log_0.1 (0.0001) = x

⇒ 0.0001 = 0.1^x

⇒ $\frac{1}{10000} = \Big(\frac{1}{10}\Big)^x$

⇒ $\frac{1}{10^4} = \Big(\frac{1}{10}\Big)^x$

⇒ $\Big(\frac{1}{10}\Big)^4 = \Big(\frac{1}{10}\Big)^x$

Equating the exponents,

⇒ x = 4.

Hence, option 1 is the correct option.

Given,

⇒ log_x 243 = 5

⇒ 243 = x⁵

⇒ 3⁵ = x⁵

⇒ x = 3.

Hence, option 2 is the correct option.

Given,

⇒ log₅ (8x - 3) = 3

⇒ (8x - 3) = 5³

⇒ 8x - 3 = 125

⇒ 8x = 125 + 3

⇒ 8x = 128

⇒ x = $\frac{128}{8}$

⇒ x = 16.

Hence, option 2 is the correct option.

Let,

⇒ log₉ 27 = x

⇒ 27 = 9^x

⇒ 3³ = (3²)^x

⇒ 3³ = 3^2x

Equating the exponents,

⇒ 2x = 3

⇒ x = $\frac{3}{2}$.

Hence, option 4 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \frac{\log \space 27}{\log \space 9} \\[1em] \Rightarrow \frac{\log \space 3^3}{\log \space 3^2} \\[1em] \Rightarrow \frac{3\log \space 3}{2\log \space 3} \\[1em] \Rightarrow \frac{3}{2}.\end{aligned}$$

Hence, option 1 is the correct option.

Given,

⇒ log_x 0.0016 = 4

⇒ 0.0016 = x⁴

⇒ (0.2)⁴ = x⁴

⇒ x = 0.2

Hence, option 2 is the correct option.

Given,

⇒ log₁₀ 8

⇒ log₁₀ 2³

⇒ 3log₁₀ 2

⇒ 3 × 0.3

⇒ 0.9

Hence, option 1 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \log_2 \space \log_{\sqrt{2}} \space \log_3 81 \\[1em] \Rightarrow \log_2 \space \log_{\sqrt{2}} \space \log_3 3^4 \\[1em] \Rightarrow \log_2 \space \log_{\sqrt{2}} \space 4\log_3 3 \\[1em] \Rightarrow \log_2 \space (\log_{\sqrt{2}} \space 4) \\[1em] \Rightarrow \log_2 \space \Big(\frac{\log \space 4}{\log \space \sqrt{2}}\Big) \\[1em] \Rightarrow \log_2 \space \Big(\frac{\log \space 2^2}{\log \space \sqrt{2}}\Big) \\[1em] \Rightarrow \log_2 \space \Big(\frac{2\log \space 2}{\frac{1}{2}\log \space 2}\Big) \\[1em] \Rightarrow \log_2 \space {\Big(\frac{2}{\frac{1}{2}}\Big)} \\[1em] \Rightarrow \log_2 \space 4 \\[1em] \Rightarrow \log_2 \space 2^2 \\[1em] \Rightarrow 2\log_2 \space 2 \\[1em] \Rightarrow 2.\end{aligned}$$

Hence, option 2 is the correct option.

Given,

⇒ log₁₀ 72

⇒ log₁₀ (9 × 8)

⇒ log₁₀ 9 + log₁₀ 8

⇒ log₁₀ 3² + log₁₀ 2 ³

⇒ 2log₁₀ 3 + 3log₁₀ 2

⇒ 2(0.4771) + 3(0.3010)

⇒ 0.9542 + 0.9030

⇒ 1.8572

Hence, option 1 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \log_x \space \frac{1}{8} = -\frac{1}{3} \\[1em] \Rightarrow x^{-\frac{1}{3}} = \frac{1}{8} \\[1em] \Rightarrow \Big(\frac{1}{x}\Big)^{\frac{1}{3}} = \Big(\frac{1}{8}\Big) \\[1em] \text{ Cubing on both sides, we get:} \\[1em] \Rightarrow \Big[\Big(\frac{1}{x}\Big)^{\frac{1}{3}}\Big]^3 = \Big(\frac{1}{8}\Big)^3 \\[1em] \Rightarrow \Big(\frac{1}{x}\Big)^{\frac{1}{3} \times 3} = \Big(\frac{1}{512}\Big) \\[1em] \Rightarrow \frac{1}{x} = \frac{1}{512} \\[1em] \Rightarrow x = 512.\end{aligned}$$

So, Assertion (A) is false.

If n^x = m, then log_n m = x.

This is the fundamental definition of a logarithm.

So, Reason (R) is true.

A is false, R is true.

Hence, option 2 is the correct option.

Given,

log (2 + 3 + 4) = log 2 + log 3 + log 4

Simplifying L.H.S,

⇒ log (2 + 3 + 4) = log (9)

Simplifying R.H.S,

⇒ log 2 + log 3 + log 4 = log (2 × 3 × 4) = log (24)

Since, L.H.S ≠ R.H.S

So, Assertion (A) is false.

⇒ log_x x = 1 and not zero.

So, Reason (R) is false.

Both A and R are false.

Hence, option 4 is the correct option.