Logarithms — Exercise 7(A)

Convert each of the following to logarithmic form:

(i) 5² = 25

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

(iii) $(64)^\dfrac{1}{3}$ = 4

(iv) 6⁰ = 1

(v) 10^-2 = 0.01

(vi) 4^-1 = $\dfrac{1}{4}$

Answer

(i) Given,

⇒ 5² = 25

⇒ log₅ (25) = 2.

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

(ii) Given,

⇒ 3^-3 = $\dfrac{1}{27}$

⇒ log₃ $\Big(\dfrac{1}{27}\Big)$ = -3.

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

(iii) Given,

⇒ $(64)^\dfrac{1}{3}$ = 4

⇒ log₆₄ 4 = $\dfrac{1}{3}$.

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

(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 = $\dfrac{1}{4}$

⇒ log₄ $\dfrac{1}{4}$ = -1.

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

Convert each of the following to exponential form:

(i) log₃ 81 = 4

(ii) log₈ 4 = $\dfrac{2}{3}$

(iii) log₂ $\dfrac{1}{8}$ = -3

(iv) log₁₀ (0.01) = -2

(v) log₅ $\Big(\dfrac{1}{5}\Big)$ = -1

(vi) log_a 1 = 0

Answer

(i) Given,

⇒ log₃ 81 = 4

⇒ 3⁴ = 81.

Hence, exponential form is 3⁴ = 81.

(ii) Given,

⇒ log₈ 4 = $\dfrac{2}{3}$

⇒ $(8)^\dfrac{2}{3}$ = 4.

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

(iii) Given,

⇒ log₂ $\dfrac{1}{8}$ = -3

⇒ 2^-3 = $\dfrac{1}{8}$.

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

(iv) Given,

⇒ log₁₀ (0.01) = -2

⇒ 10^-2 = 0.01.

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

(v) Given,

⇒ log₅ $\Big(\dfrac{1}{5}\Big)$ = -1

⇒ 5^-1 = $\Big(\dfrac{1}{5}\Big)$.

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

(vi) Given,

⇒ log_a 1 = 0

⇒ a⁰ = 1.

Hence, exponential form is a⁰ = 1.

By converting to exponential form, find the value of each of the following:

(i) log₂ 64

(ii) log₈ 32

(iii) log₃ $\dfrac{1}{9}$

(iv) log_0.5 (16)

(v) log₂ (0.125)

(vi) log₇ 7

Answer

(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 = $\dfrac{5}{3}$.

Hence, log₈ 32 = $\dfrac{5}{3}$.

(iii) Let,

$\Rightarrow \log_{3} \space {\dfrac{1}{9}} = x \\[1em] \Rightarrow \dfrac{1}{9} = 3^x \\[1em] \Rightarrow \dfrac{1}{3^2} = 3^x \\[1em] \Rightarrow 3^{-2} = 3^x \\[1em]$

Equating the exponents,

⇒ x = -2.

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

(iv) Let,

⇒ log_0.5 (16) = x

⇒ 16 = 0.5^x

⇒ 2⁴ = $\Big(\dfrac{1}{2}\Big)^x$

⇒ 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

⇒ $\dfrac{125}{1000}$ = 2^x

⇒ $\dfrac{1}{8}$ = 2^x

⇒ $\dfrac{1}{2^3}$ = 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.

Find the value of x, when:

(i) log₂ x = -2

(ii) log_x 9 = 1

(iii) log₉ 243 = x

(iv) log₃ x = 0

(v) $\log _{\sqrt{3}}$ (x − 1) = 2

(vi) log₅ (x² − 19) = 3

(vii) log_x 64 = $\dfrac{3}{2}$

(viii) log₂ (x² − 9) = 4

(ix) log_x (0.008) = −3

Answer

(i) Given,

⇒ log₂ x = -2

⇒ x = 2^-2

⇒ x = $\dfrac{1}{2^2}$

⇒ x = $\dfrac{1}{4}$

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

(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 = $\dfrac{5}{2}$.

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

(iv) Given,

⇒ log₃ x = 0

⇒ x = 3⁰

⇒ x = 1.

Hence, x = 1.

(v) Given,

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

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,

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

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

⇒ $\dfrac{8}{1000}$ = x⁻³

⇒ $\dfrac{1}{125}$ = x⁻³

⇒ $\dfrac{1}{5^3}$ = x⁻³

⇒ 5⁻³ = x⁻³

Equating the bases,

⇒ x = 5.

Hence, x = 5.

If log₁₀ x = p and log₁₀ y = q, show that xy = (10)^{p + q}.

Answer

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, log₁₀ y = b,

(i) Write down 10^{a + 1} in terms of x.

(ii) Write down 10^2b in terms of y.

(iii) If log₁₀ P = 2a − b, express P in terms of x and y.

Answer

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 = $\dfrac{x^2}{y}$.

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