Logarithms — Exercise 8(B)

The value of 3 + log₅ 5^-2

1. 1

2. 5

3. 3 - $\dfrac{1}{5}$

4. 3 + $\dfrac{1}{5}$

Answer

Given,

⇒ 3 + log₅ 5^-2

⇒ 3 + (-2log₅ 5)

⇒ 3 + (-2 × 1)

⇒ 3 - 2

⇒ 1.

Hence, Option 1 is the correct option.

The value of log₅ 75 - log₅ 3 is :

1. 72

2. 2

3. 25

4. 5

Answer

Given,

$\Rightarrow \text{log}_{5} \space 75 - \text{log}_{5} \space 3 \\[1em] \Rightarrow \text{log}_{5} \space {\dfrac{75}{3}} \\[1em] \Rightarrow \text{log}_{5} \space 25 \\[1em] \Rightarrow \text{log}_{5} \space 5^2 \\[1em] \Rightarrow 2 \times \text{log}_{5} \space 5 \\[1em] \Rightarrow 2 \times 1 \\[1em] \Rightarrow 2.$

Hence, Option 2 is the correct option.

The value of log₅ 125 ÷ log₅ $\sqrt{5}$ is :

1. 5

2. 120

3. 6

4. 60

Answer

Given,

$\Rightarrow \text{log}_{5} \space 125 ÷ \text{log}_{5} \space \sqrt{5} \\[1em] \Rightarrow \text{log}_{5} \space 5^3 ÷ \text{log}_{5} \space (5)^{\dfrac{1}{2}} \\[1em] \Rightarrow 3 \times \text{log}_{5} \space 5 ÷ \dfrac{1}{2} \times \text{log}_{5} \space 5 \\[1em] \Rightarrow 3 \times 1 ÷ \dfrac{1}{2} \times 1 \\[1em] \Rightarrow 3 ÷ \dfrac{1}{2} \\[1em] \Rightarrow 3 \times 2 \\[1em] \Rightarrow 6.$

Hence, Option 3 is the correct option.

The value of $(\sqrt{x})^{\text{4log}_{x} \space a}$ is :

1. a

2. ax

3. $\dfrac{1}{2}ax$

4. a²

Answer

Given,

$\Rightarrow (\sqrt{x})^{\text{4log}_{x} \space a} \\[1em] \Rightarrow (x^{\dfrac{1}{2}})^{\text{4log}_{x} \space a} \\[1em] \Rightarrow (x)^{\dfrac{1}{2} \times \text{4log}_{x} \space a} \\[1em] \Rightarrow (x)^{\text{2log}_{x} \space a} \\[1em] \Rightarrow (x)^{\text{log}_{x} \space a^2} \\[1em] \Rightarrow a^2.$

Hence, Option 4 is the correct option.

If $\dfrac{\text{log }125}{\text{log } \dfrac{1}{5}}$ = log x, the value of x is :

1. 0.001

2. 0.01

3. 25

4. 5

Answer

Given,

$\Rightarrow \dfrac{\text{log }125}{\text{log } \dfrac{1}{5}} = \text{log x} \\[1em] \Rightarrow \dfrac{\text{log }5^3}{\text{log } 5^{-1}} = \text{log x} \\[1em] \Rightarrow \dfrac{3\text{log }5}{-1\text{log } 5} = \text{log x} \\[1em] \Rightarrow -3 = \text{log x} \\[1em] \Rightarrow x = 10^{-3} \\[1em] \Rightarrow x = \dfrac{1}{10^3} \\[1em] \Rightarrow x = \dfrac{1}{1000} \\[1em] \Rightarrow x = 0.001$

Hence, Option 1 is the correct option.

If log (x - 5) + log (x + 5) = 2 log 12, the positive value of x is :

1. 5

2. 13

3. 4.8

4. 12

Answer

Given,

⇒ log (x - 5) + log (x + 5) = 2 log 12

⇒ log (x - 5)(x + 5) = log 12²

⇒ log (x² + 5x - 5x - 25) = log 144

⇒ log (x² - 25) = log 144

⇒ x² - 25 = 144

⇒ x² = 144 + 25

⇒ x² = 169

⇒ x = $\sqrt{169}$

⇒ x = $\pm 13$.

So, positive value of x = 13.

Hence, Option 2 is the correct option.

Express in terms of log 2 and log 3 :

log 36

Answer

Simplifying the expression,

⇒ log 36

⇒ log (2² × 3²)

⇒ log 2² + log 3²

⇒ 2 log 2 + 2 log 3.

Hence, log 36 = 2 log 2 + 2 log 3.

Express in terms of log 2 and log 3 :

log 144

Answer

Simplifying the expression,

⇒ log 144

⇒ log (2⁴ × 3²)

⇒ log 2⁴ + log 3²

⇒ 4 log 2 + 2 log 3.

Hence, log 144 = 4 log 2 + 2 log 3.

Express in terms of log 2 and log 3 :

log 4.5

Answer

Simplifying the expression,

⇒ log 4.5

⇒ log $\dfrac{45}{10}$

⇒ log $\dfrac{9}{2}$

⇒ log 9 - log 2

⇒ log 3² - log 2

⇒ 2 log 3 - log 2.

Hence, log 4.5 = 2 log 3 - log 2.

Express in terms of log 2 and log 3 :

$\text{log } \dfrac{26}{51} - \text{log } \dfrac{91}{119}$

Answer

Simplifying the expression,

$\Rightarrow \text{log } \dfrac{26}{51} - \text{log } \dfrac{91}{119} \\[1em] \Rightarrow \text{log } \Big(\dfrac{26}{51} ÷ \dfrac{91}{119}\Big) \\[1em] \Rightarrow \text{log } \Big(\dfrac{26}{51} \times \dfrac{119}{91}\Big) \\[1em] \Rightarrow \text{log } \dfrac{3094}{4641} \\[1em] \Rightarrow \text{log } \dfrac{2}{3} \\[1em] \Rightarrow \text{log } 2 - \text{log 3}.$

Hence, $\text{log } \dfrac{26}{51} - \text{log } \dfrac{91}{119} = \text{log 2 - log 3}$.

Express in terms of log 2 and log 3 :

$\text{log } \dfrac{75}{16} - \text{2 log } \dfrac{5}{9} + \text{log } \dfrac{32}{243}$

Answer

Simplifying the expression,

$\Rightarrow \text{log } \dfrac{75}{16} - \text{2 log } \dfrac{5}{9} + \text{log } \dfrac{32}{243} \\[1em]$

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

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

⇒ log 75 - log 16 - log 5² + log 9² + log 32 - log 243

⇒ log 75 - log 16 - log 25 + log 81 + log 32 - log 243

⇒ log 75 + log 81 + log 32 - log 16 - log 25 - log 243

⇒ log 75 + log 81 + log 32 - (log 16 + log 25 + log 243)

⇒ log (75 × 81 × 32) - log (16 × 25 × 243)

⇒ log $\dfrac{75 \times 81 \times 32}{16 \times 25 \times 243}$

⇒ log 2.

Hence, $\text{log } \dfrac{75}{16} - \text{2 log } \dfrac{5}{9} + \text{log } \dfrac{32}{243}$ = log 2.

Express the following in a form free from logarithm :

2 log x - log y = 1

Answer

Given,

⇒ 2 log x - log y = 1

⇒ log x² - log y = 1

⇒ log $\dfrac{x^2}{y}$ = 1

⇒ $\dfrac{x^2}{y}$ = 10¹

⇒ x² = 10y.

Hence, required equation is x² = 10y.

Express the following in a form free from logarithm :

2 log x + 3 log y = log a

Answer

Given,

⇒ 2 log x + 3 log y = log a

⇒ log x² + log y³ = log a

⇒ log (x².y³) = log a

⇒ x².y³ = a

Hence, required equation is x²y³ = a.

Express the following in a form free from logarithm :

a log x - b log y = 2 log 3

Answer

Given,

⇒ a log x - b log y = 2 log 3

⇒ log x^a - log y^b = log 3²

⇒ log $\Big(\dfrac{x^a}{y^b}\Big)$ = log 3²

⇒ log $\Big(\dfrac{x^a}{y^b}\Big)$ = log 9

⇒ $\Big(\dfrac{x^a}{y^b}\Big) = 9$

⇒ x^a = 9y^b.

Hence, required equation is x^a = 9y^b.

Evaluate :

log 5 + log 8 - 2 log 2

Answer

Evaluating the expression,

⇒ log 5 + log 8 - 2 log 2

⇒ log 5 + log 8 - log 2²

⇒ log $\dfrac{5 \times 8}{2^2}$

⇒ log $\dfrac{40}{4}$

⇒ log 10

⇒ 1.

Hence, log 5 + log 8 - 2 log 2 = 1.

Evaluate :

log₁₀ 8 + log₁₀ 25 + 2 log₁₀ 3 - log₁₀ 18

Answer

Evaluating the expression,

⇒ log₁₀ 8 + log₁₀ 25 + 2 log₁₀ 3 - log₁₀ 18

⇒ log₁₀ 8 + log₁₀ 25 + log₁₀ 3² - log₁₀ 18

⇒ log₁₀ $\dfrac{8 \times 25 \times 3^2}{18}$

⇒ log₁₀ $\dfrac{1800}{18}$

⇒ log₁₀ 100

⇒ log₁₀ 10²

⇒ 2 × log₁₀ 10

⇒ 2 × 1

⇒ 2.

Hence, log₁₀ 8 + log₁₀ 25 + 2 log₁₀ 3 - log₁₀ 18 = 2.

Evaluate :

$\text{log 4} + \dfrac{1}{3}\text{ log 125} - \dfrac{1}{5}\text{ log 32}$

Answer

Evaluating the expression,

$\Rightarrow \text{log 4} + \dfrac{1}{3}\text{ log 125} - \dfrac{1}{5}\text{ log 32} \\[1em] \Rightarrow \text{log 4} + \dfrac{1}{3}\text{ log 5}^3 - \dfrac{1}{5}\text{ log 2}^5 \\[1em] \Rightarrow \text{log 4 + log (5)}^{\Big(3 \times \dfrac{1}{3}\Big)} - \text{log (2)}^{\Big(5 \times \dfrac{1}{5}\Big)} \\[1em] \Rightarrow \text{log 4 + log 5 - log 2} \\[1em] \Rightarrow \text{log } \dfrac{4 \times 5}{2} \\[1em] \Rightarrow \text{log } \dfrac{20}{2} \\[1em] \Rightarrow \text{log } 10 \\[1em] \Rightarrow 1.$

Hence, $\text{log 4} + \dfrac{1}{3}\text{ log 125} - \dfrac{1}{5}\text{ log 32}$ = 1.

Prove that :

$\text{2 log }\dfrac{15}{18} - \text{log } \dfrac{25}{162} + \text{log } \dfrac{4}{9} = \text{log 2}$

Answer

To prove:

$\text{2 log }\dfrac{15}{18} - \text{log } \dfrac{25}{162} + \text{log } \dfrac{4}{9} = \text{log 2}$

Solving L.H.S. of the equation, we get :

$\Rightarrow \text{2 log }\dfrac{15}{18} - \text{log } \dfrac{25}{162} + \text{log } \dfrac{4}{9} \\[1em] \Rightarrow \text{log } \Big(\dfrac{15}{18}\Big)^2 + \text{log } \dfrac{4}{9} - \text{log } \dfrac{25}{162} \\[1em] \Rightarrow \text{log } \dfrac{225}{324} + \text{log } \dfrac{4}{9} - \text{log } \dfrac{25}{162} \\[1em] \Rightarrow \text{log } \dfrac{\dfrac{225}{324} \times \dfrac{4}{9}}{\dfrac{25}{162}} \\[1em] \Rightarrow \text{log } \dfrac{\dfrac{900}{2916}}{\dfrac{25}{162}} \\[1em] \Rightarrow \text{log } \dfrac{900 \times 162}{25 \times 2916} \\[1em] \Rightarrow \text{log } \dfrac{145800}{72900} \\[1em] \Rightarrow \text{log } 2.$

Since, L.H.S. = R.H.S.

Hence, proved that $\text{2 log }\dfrac{15}{18} - \text{log } \dfrac{25}{162} + \text{log } \dfrac{4}{9} = \text{log 2}$.

Find x, if :

x - log 48 + 3 log 2 = $\dfrac{1}{3}$ log 125 - log 3

Answer

Given,

⇒ x - log 48 + 3 log 2 = $\dfrac{1}{3}$ log 125 - log 3

⇒ x - log 48 + log 2³ = log $(125)^{\dfrac{1}{3}}$ - log 3

⇒ x - log 48 + log 8 = log $(5^3)^{\dfrac{1}{3}}$ - log 3

⇒ x - log 48 + log 8 = log 5 - log 3

⇒ x = log 5 + log 48 - log 3 - log 8

⇒ x = (log 5 + log 48) - (log 3 + log 8)

⇒ x = log (5 × 48) - log (3 × 8)

⇒ x = log 240 - log 24

⇒ x = log $\dfrac{240}{24}$

⇒ x = log 10

⇒ x = 1.

Hence, x = 1.

Express log₁₀ 2 + 1 in the form of log₁₀ x.

Answer

Given,

⇒ log₁₀ 2 + 1

⇒ log₁₀ 2 + log₁₀ 10

⇒ log₁₀ (2 × 10)

⇒ log₁₀ 20.

Hence, log₁₀ 2 + 1 = log₁₀ 20.

Solve for x :

log₁₀ (x - 10) = 1

Answer

Given,

⇒ log₁₀ (x - 10) = 1

⇒ x - 10 = 10¹

⇒ x - 10 = 10

⇒ x = 10 + 10 = 20.

Hence, x = 20.

Solve for x :

log (x² - 21) = 2

Answer

Given,

⇒ log (x² - 21) = 2

⇒ x² - 21 = 10²

⇒ x² - 21 = 100

⇒ x² = 100 + 21

⇒ x² = 121

⇒ x = $\sqrt{121}$

⇒ x = $\pm 11$.

Hence, x = $\pm 11$.

Solve for x :

log (x - 2) + log (x + 2) = log 5

Answer

Given,

⇒ log (x - 2) + log (x + 2) = log 5

⇒ log (x - 2)(x + 2) = log 5

⇒ log (x² + 2x - 2x - 4) = log 5

⇒ log (x² - 4) = log 5

⇒ x² - 4 = 5

⇒ x² = 5 + 4

⇒ x² = 9

⇒ x = $\sqrt{9} = \pm 3$.

Since, x cannot be negative as that will make (x - 2) and (x + 2) negative.

Hence, x = 3.

Solve for x :

log (x + 5) + log (x - 5) = 4 log 2 + 2 log 3

Answer

Given,

⇒ log (x + 5) + log (x - 5) = 4 log 2 + 2 log 3

⇒ log (x + 5)(x - 5) = log 2⁴ + log 3²

⇒ log (x² - 5x + 5x - 25) = log (2⁴ × 3²)

⇒ log (x² - 25) = log (16 × 9)

⇒ log (x² - 25) = log 144

⇒ x² - 25 = 144

⇒ x² = 144 + 25

⇒ x² = 169

⇒ x = $\sqrt{169} = \pm 13$.

Since, x cannot be negative as that will make (x + 5) and (x - 5) negative.

Hence, x = 13.

Solve for x :

$\dfrac{\text{log 81}}{\text{log 27}}$ = x

Answer

Given,

$\Rightarrow \dfrac{\text{log 81}}{\text{log 27}} = x \\[1em] \Rightarrow \dfrac{\text{log 3}^4}{\text{log 3}^3} = x \\[1em] \Rightarrow \dfrac{\text{4 log 3}}{\text{3 log 3}} = x \\[1em] \Rightarrow x = \dfrac{4}{3} = 1\dfrac{1}{3}.$

Hence, x = $1\dfrac{1}{3}$.

Solve for x :

$\dfrac{\text{log 128}}{\text{log 32}}$ = x

Answer

Given,

$\Rightarrow \dfrac{\text{log 128}}{\text{log 32}} = x \\[1em] \Rightarrow \dfrac{\text{log 2}^7}{\text{log 2}^5} = x \\[1em] \Rightarrow \dfrac{\text{7 log 2}}{\text{5 log 2}} = x \\[1em] \Rightarrow x = \dfrac{7}{5} = 1.4$

Hence, x = 1.4

Solve for x :

$\dfrac{\text{log 64}}{\text{log 8}}$ = log x

Answer

Given,

$\Rightarrow \dfrac{\text{log 64}}{\text{log 8}} = \text{log x} \\[1em] \Rightarrow \dfrac{\text{log 2}^6}{\text{log 2}^3} = \text{log x} \\[1em] \Rightarrow \dfrac{\text{6 log 2}}{\text{3 log 2}} = \text{log x} \\[1em] \Rightarrow \text{log x} = 2 \\[1em] \Rightarrow x = 10^2 = 100.$

Hence, x = 100.

Solve for x :

$\dfrac{\text{log 225}}{\text{log 15}}$ = log x

Answer

Given,

$\Rightarrow \dfrac{\text{log 225}}{\text{log 15}} = \text{log x} \\[1em] \Rightarrow \dfrac{\text{log 15}^2}{\text{log 15}} = \text{log x} \\[1em] \Rightarrow \dfrac{\text{2 log 15}}{\text{log 15}} = \text{log x} \\[1em] \Rightarrow \text{log x} = 2 \\[1em] \Rightarrow x = 10^2 = 100.$

Hence, x = 100.

Given log x = m + n and log y = m - n, express the value of log $\dfrac{10x}{y^2}$ in terms of m and n.

Answer

Given,

⇒ log x = m + n

⇒ x = 10^{m + n} ........(1)

Given,

⇒ log y = m - n

⇒ y = 10^{m - n} ........(2)

Substituting value of x and y from equation (1) and equation (2) in log $\dfrac{10x}{y^2}$, we get :

$\Rightarrow \text{log} \dfrac{10x}{y^2} = \text{log } \dfrac{10 \times 10^{m + n}}{(10^{m - n})^2} \\[1em] = \text{log } \dfrac{10^{m + n + 1}}{10^{2(m - n)}} \\[1em] = \text{log } 10^{m + n + 1 - 2(m - n)} \\[1em] = \text{log } 10^{m + n + 1 - 2m + 2n} \\[1em] = \text{log } 10^{3n - m + 1} \\[1em] = (3n - m + 1) \text{ log} 10 \\[1em] = (3n - m + 1) \times 1 \\[1em] = 3n - m + 1 \\[1em] = 1 - m + 3n$

Hence, log $\dfrac{10x}{y^2}$ = 1 - m + 3n.

State, true or false :

log 1 × log 1000 = 0

Answer

Given,

log 1 × log 1000 = 0

Solving L.H.S. of the above equation, we get :

⇒ log 1 × log 1000

⇒ 0 × log 1000

⇒ 0.

Since, L.H.S. = R.H.S.

Hence, the statement "log 1 × log 1000 = 0" is true.

State, true or false :

$\dfrac{\text{log x}}{\text{log y}}$ = log x - log y

Answer

Given,

$\dfrac{\text{log x}}{\text{log y}}$ = log x - log y

Solving R.H.S. of the above equation, we get :

⇒ log x - log y

⇒ log $\dfrac{x}{y}$.

Since, L.H.S. ≠ R.H.S.

Hence, the statement $\dfrac{\text{log x}}{\text{log y}} = \text{log x} - \text{log y}$ is false.

State, true or false :

If $\dfrac{\text{log 25}}{\text{log 5}}$ = log x, then x = 2

Answer

Given,

$\dfrac{\text{log 25}}{\text{log 5}}$ = log x

Solving the equation, we get :

$\Rightarrow \dfrac{\text{log 25}}{\text{log 5}} = \text{log x} \\[1em] \Rightarrow \dfrac{\text{log 5}^2}{\text{log 5}} = \text{log x} \\[1em] \Rightarrow \dfrac{\text{2 log 5}}{\text{log 5}} = \text{log x} \\[1em] \Rightarrow \text{log x} = 2 \\[1em] \Rightarrow x = 10^2 = 100.$

Since, x is not equal to 2.

Hence, the statement $\dfrac{\text{log 25}}{\text{log 5}}$ = log x, then x = 2 is false.

State, true or false :

log x × log y = log x + log y

Answer

Given,

log x × log y = log x + log y

Solving the R.H.S. of the above equation, we get :

⇒ log x + log y

⇒ log xy

Since, L.H.S. ≠ R.H.S.

Hence, the statement log x × log y = log x + log y is false.

If log₁₀ 2 = a and log₁₀ 3 = b; express log 12 in terms of 'a' and 'b'.

Answer

Given,

log₁₀2 = a and log₁₀3 = b

Simplifying the expression :

⇒ log 12

⇒ log (2² × 3)

⇒ log 2² + log 3

⇒ 2 log 2 + log 3

⇒ 2a + b.

Hence, log 12 = 2a + b.

If log₁₀ 2 = a and log₁₀ 3 = b; express log 2.25 in terms of 'a' and 'b'.

Answer

Given,

log₁₀2 = a and log₁₀3 = b

Simplifying the expression :

⇒ log 2.25

⇒ log $\dfrac{225}{100}$

⇒ log $\dfrac{9}{4}$

⇒ log 9 - log 4

⇒ log 3² - log 2²

⇒ 2 log 3 - 2 log 2

⇒ 2b - 2a.

Hence, log 2.25 = 2b - 2a.

If log₁₀ 2 = a and log₁₀ 3 = b; express log $2\dfrac{1}{4}$ in terms of 'a' and 'b'.

Answer

Given,

log₁₀ 2 = a and log₁₀ 3 = b

Simplifying the expression :

$\Rightarrow \text{log } 2\dfrac{1}{4} \\[1em] \Rightarrow \text{log } \dfrac{9}{4} \\[1em] \Rightarrow \text{log 9 - log 4} \\[1em] \Rightarrow \text{log 3}^2 - \text{log 2}^2 \\[1em] \Rightarrow \text{2 log 3 - 2 log 2} \\[1em] \Rightarrow \text{2b - 2a}.$

Hence, $\text{log } 2\dfrac{1}{4}$ = 2b - 2a.

If log₁₀ 2 = a and log₁₀ 3 = b; express log 5.4 in terms of 'a' and 'b'.

Answer

Given,

log₁₀ 2 = a and log₁₀ 3 = b

Simplifying the expression :

$\Rightarrow \text{log } 5.4 \\[1em] \Rightarrow \text{log } \dfrac{54}{10} \\[1em] \Rightarrow \text{log 54 - log 10} \\[1em] \Rightarrow \text{log }(2 \times 3^3) - 1 \\[1em] \Rightarrow \text{log 2} + \text{log }3^3 - 1 \\[1em] \Rightarrow \text{log 2 + 3 log 3} - 1 \\[1em] \Rightarrow a + 3b - 1.$

Hence, log 5.4 = a + 3b - 1.

If log₁₀ 2 = a and log₁₀ 3 = b; express log 60 in terms of 'a' and 'b'.

Answer

Given,

log₁₀ 2 = a and log₁₀ 3 = b

Simplifying the expression :

⇒ log 60

⇒ log (2 × 3 × 10)

⇒ log 2 + log 3 + log 10

⇒ a + b + 1.

Hence, log 60 = a + b + 1.

If log₁₀ 2 = a and log₁₀ 3 = b; express log $3\dfrac{1}{8}$ in terms of 'a' and 'b'.

Answer

Given,

log₁₀ 2 = a and log₁₀ 3 = b

Simplifying the expression :

$\Rightarrow \text{log } 3\dfrac{1}{8} \\[1em] \Rightarrow \text{log } \dfrac{25}{8} \\[1em] \Rightarrow \text{log 25 - log 8} \\[1em] \Rightarrow \text{log 5}^2 - \text{log 2}^3 \\[1em] \Rightarrow \text{2 log 5 - 3 log 2} \\[1em] \Rightarrow \text{2 log } \dfrac{10}{2} - \text{3 log 2} \\[1em] \Rightarrow \text{2 (log 10 - log 2)} - \text{3 log 2} \\[1em] \Rightarrow \text{2 log 10 - 2 log 2 - 3 log 2} \\[1em] \Rightarrow 2 \times 1 - \text{5 log 2} \\[1em] \Rightarrow 2 - 5a.$

Hence, $\text{log } 3\dfrac{1}{8}$ = 2 - 5a.

If log 2 = 0.3010 and log 3 = 0.4771; find the value of log 12.

Answer

Simplifying the expression,

⇒ log 12

⇒ log (2² × 3)

⇒ log 2² + log 3

⇒ 2 log 2 + log 3

⇒ 2 × 0.3010 + 0.4771

⇒ 0.6020 + 0.4771

⇒ 1.0791

Hence, log 12 = 1.0791

If log 2 = 0.3010 and log 3 = 0.4771; find the value of log 1.2.

Answer

Simplifying the expression,

⇒ log 1.2

⇒ log $\dfrac{12}{10}$

⇒ log 12 - log 10

⇒ log (2² × 3) - log 10

⇒ log 2² + log 3 - log 10

⇒ 2 log 2 + log 3 - 1

⇒ 2 × 0.3010 + 0.4771 - 1

⇒ 0.6020 + 0.4771 - 1

⇒ 1.0791 - 1

⇒ 0.0791

Hence, log 1.2 = 0.0791

If log 2 = 0.3010 and log 3 = 0.4771; find the value of log 3.6.

Answer

Simplifying the expression,

⇒ log 3.6

⇒ log $\dfrac{36}{10}$

⇒ log 36 - log 10

⇒ log (12 × 3) - log 10

⇒ log 12 + log 3 - log 10

⇒ log (2² × 3) + log 3 - log 10

⇒ log 2² + log 3 + log 3 - log 10

⇒ 2 log 2 + 2 log 3 - 1

⇒ 2 × 0.3010 + 2 × 0.4771 - 1

⇒ 0.6020 + 0.9542 - 1

⇒ 1.5562 - 1

⇒ 0.5562

Hence, log 3.6 = 0.5562

If log 2 = 0.3010 and log 3 = 0.4771; find the value of log 15.

Answer

Simplifying the expression,

⇒ log 15

⇒ log (3 × 5)

⇒ log 3 + log 5

⇒ log 3 + log $\dfrac{10}{2}$

⇒ log 3 + log 10 - log 2

⇒ 0.4771 + 1 - 0.3010

⇒ 1.1761

Hence, log 15 = 1.1761

If log 2 = 0.3010 and log 3 = 0.4771; find the value of log 25.

Answer

Simplifying the expression,

⇒ log 25

⇒ log 5²

⇒ 2 log 5

⇒ 2 log $\dfrac{10}{2}$

⇒ 2(log 10 - log 2)

⇒ 2(1 - 0.3010)

⇒ 2 × 0.699

⇒ 1.3980

Hence, log 25 = 1.3980

If log 2 = 0.3010 and log 3 = 0.4771; find the value of $\dfrac{2}{3}$ log 8.

Answer

Simplifying the expression,

$\Rightarrow \dfrac{2}{3} \text{ log 8} \\[1em] \Rightarrow \dfrac{2}{3} \text{ log 2}^3 \\[1em] \Rightarrow \dfrac{2}{3} \times 3 \text{ log 2} \\[1em] \Rightarrow 2 \text{log 2} \\[1em] \Rightarrow 2 \times 0.3010 \\[1em] \Rightarrow 0.6020$

Hence, $\dfrac{2}{3}$ log 8 = 0.6020

Given 2 log₁₀ x + 1 = log₁₀ 250, find :

(i) x

(ii) log₁₀ 2x

Answer

(i) Given,

⇒ 2 log₁₀ x + 1 = log₁₀ 250

⇒ log₁₀ x² + log₁₀ 10 = log₁₀ 250

⇒ log₁₀ (x² × 10) = log₁₀ 250

⇒ log₁₀ (10x²) = log₁₀ 250

⇒ 10x² = 250

⇒ x² = $\dfrac{250}{10}$

⇒ x² = 25

⇒ x = $\sqrt{25}$ = 5.

Hence, x = 5.

(ii) Substituting value of x in log₁₀ 2x, we get :

⇒ log₁₀ 2x

⇒ log₁₀ 2(5)

⇒ log₁₀ 10

⇒ 1.

Hence, log₁₀ 2x = 1.

Given 3 log x + $\dfrac{1}{2}$ log y = 2, express y in terms of x.

Answer

Given,

$\Rightarrow \text{3 log x} + \dfrac{1}{2} \text{log y} = 2 \\[1em] \Rightarrow \text{log } x^3 + \text{log } y^{\dfrac{1}{2}} = 2 \\[1em] \Rightarrow \text{log } x^3\sqrt{y} = 2 \\[1em] \Rightarrow x^3\sqrt{y} = 10^2$

Squaring both sides, we get :

$\Rightarrow (x^3\sqrt{y})^2 = (10^2)^2 \\[1em] \Rightarrow x^6y = 10^4 \\[1em] \Rightarrow y = \dfrac{10^4}{x^6} = 10000x^{-6}.$

Hence, y = 10000x^-6.