Logarithms — Exercise 8.1

Convert the following to logarithmic form:

(i) 5² = 25

(ii) a⁵ = 64

(iii) 7^x = 100

(iv) 9⁰ = 1

(v) 6¹ = 6

(vi) 3^-2 = $\dfrac{1}{9}$

(vii) 10^-2 = 0.01

(viii) $(81)^{\dfrac{3}{4}}$ = 27.

Answer

(i) 5² = 25

⇒ log₅ 25 = 2.

(ii) a⁵ = 64

⇒ log_a 64 = 5.

(iii) 7^x = 100

⇒ log₇ 100 = x.

(iv) 9⁰ = 1

⇒ log₉ 1 = 0.

(v) 6¹ = 6

⇒ log₆ 6 = 1.

(vi) 3^-2 = $\dfrac{1}{9}$

⇒ log₃ $\dfrac{1}{9}$ = -2.

(vii) 10^-2 = 0.01

⇒ log₁₀ 0.01 = -2.

(viii) $(81)^{\dfrac{3}{4}}$ = 27

⇒ log₈₁ 27 = $\dfrac{3}{4}$

Convert the following into exponential form:

(i) log₂32 = 5

(ii) log₃81 = 4

(iii) log₃$\dfrac{1}{3}$ = -1

(iv) log₈4 = $\dfrac{2}{3}$

(v) log₈32 = $\dfrac{5}{3}$

(vi) log₁₀ (0.001) = -3

(vii) log₂ 0.25 = -2

(viii) log_a $\Big(\dfrac{1}{a}\Big)$ = -1

Answer

(i) log₂32 = 5

⇒ 2⁵ = 32

(ii) log₃81 = 4

⇒ 3⁴ = 81

(iii) log₃$\dfrac{1}{3}$ = -1

⇒ 3^-1 = $\dfrac{1}{3}$

(iv) log₈4 = $\dfrac{2}{3}$

⇒ $(8)^{\dfrac{2}{3}}$ = 4

(v) log₈ 32 = $\dfrac{5}{3}$

⇒ $(8)^{\dfrac{5}{3}} = 32$

(vi) log₁₀ (0.001) = -3

⇒ 10^-3 = 0.001

(vii) log₂ 0.25 = -2

⇒ (2)^-2 = 0.25

(viii) log_a $\Big(\dfrac{1}{a}\Big)$ = -1

⇒ a^-1 = $\dfrac{1}{a}$.

By converting to exponential form, find the values of :

(i) log₂16

(ii) log₅125

(iii) log₄8

(iv) log₉27

(v) log₁₀ (0.01)

(vi) log₇ $\dfrac{1}{7}$

(vii) log_0.5 256

(viii) log₂ 0.25

Answer

(i) log₂16 = x

⇒ 2^x = 16

⇒ 2^x = 2⁴

∴ x = 4.

Hence, log₂16 = 4.

(ii) log₅125 = x

⇒ 5^x = 125

⇒ 5^x = 5³

∴ x = 3.

Hence, log₅125 = 3.

(iii) log₄8 = x

⇒ 4^x = 8

⇒ (2²)^x = 2³

⇒ 2^2x = 2³

⇒ 2x = 3

⇒ x = $\dfrac{3}{2}.$

Hence, log₄8 = $\dfrac{3}{2}$.

(iv) log₉27 = x

⇒ 9^x = 27

⇒ (3²)^x = 3³

⇒ 3^2x = 3³

⇒ 2x = 3

⇒ x = $\dfrac{3}{2}.$

Hence, log₉27 = $\dfrac{3}{2}$.

(v) log₁₀ (0.01) = x

⇒ 10^x = 0.01

⇒ 10^x = 10^-2

∴ x = -2.

Hence, log₁₀ (0.01) = -2.

(vi) log₇$\dfrac{1}{7}$ = x

⇒ 7^x = $\dfrac{1}{7}$

⇒ 7^x = 7^-1

∴ x = -1.

Hence, log₇$\dfrac{1}{7}$ = -1.

(vii) log_0.5 256 = x

⇒ (0.5)^x = 256

⇒ $\Big(\dfrac{5}{10}\Big)^x$ = (2)⁸

⇒ $\Big(\dfrac{1}{2}\Big)^x$ = (2)⁸

⇒ (2)^-x = (2)⁸

∴ -x = 8 ⇒ x = -8.

Hence, log_0.5256 = -8.

(viii) log₂ 0.25 = x

⇒ 2^x = 0.25

⇒ 2^x = $\dfrac{25}{100}$

⇒ 2^x = $\dfrac{1}{4} = \dfrac{1}{2^2}$

⇒ 2^x = 2^-2

⇒ x = -2.

Hence, log₂ 0.25 = -2.

Solve the following equation for x:

log₃x = 2

Answer

Given,

⇒ log₃x = 2

⇒ x = 3² = 9.

Hence, x = 9.

Solve the following equation for x:

log_x25 = 2

Answer

Given,

⇒ log_x25 = 2

⇒ 25 = x²

⇒ 5² = x²

⇒ x = 5.

Hence, x = 5.

Solve the following equation for x:

log₁₀x = -2

Answer

Given,

⇒ log₁₀x = -2

⇒ x = 10^-2

⇒ x = $\dfrac{1}{10^2} = \dfrac{1}{100}$ = 0.01.

Hence, x = 0.01.

Solve the following equation for x:

log₄x = $\dfrac{1}{2}$

Answer

Given,

⇒ log₄x = $\dfrac{1}{2}$

⇒ x = $4^{\dfrac{1}{2}} = \sqrt{4}$ = 2.

Hence, x = 2.

Solve the following equation for x:

log_x11 = 1

Answer

Given,

⇒ log_x11 = 1

⇒ x¹ = 11

⇒ x = 11.

Hence, x = 11.

Solve the following equation for x:

log_x$\dfrac{1}{4}$ = -1

Answer

Given,

⇒ log_x$\dfrac{1}{4}$ = -1

$\Rightarrow x^{-1} = \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{1}{x} = \dfrac{1}{4} \\[1em] \Rightarrow x = 4.$

Hence, x = 4.

Solve the following equation for x:

log₈₁x = $\dfrac{3}{2}$

Answer

Given,

⇒ log₈₁x = $\dfrac{3}{2}$

$\Rightarrow x = (81)^{\dfrac{3}{2}} \\[1em] \Rightarrow x = (9)^2{\dfrac{3}{2}} \\[1em] \Rightarrow x = 9^3 = 729.$

Hence, x = 729.

Solve the following equation for x:

log₉x = 2.5

Answer

Given,

⇒ log₉x = 2.5

⇒ x = 9^2.5

⇒ x = $(3^2)^{\dfrac{5}{2}} = 3^{2 \times \dfrac{5}{2}}$

⇒ x = 3⁵ = 243.

Hence, x = 243.

Solve the following equation for x:

log₄x = -1.5

Answer

Given,

⇒ log₄x = -1.5

⇒ x = 4^-1.5

⇒ x = $4^{-\dfrac{3}{2}} = (2^2)^{-\dfrac{3}{2}}$

⇒ x = 2^-3 = $\dfrac{1}{2^3} = \dfrac{1}{8}$.

Hence, x = $\dfrac{1}{8}$.

Solve the following equation for x:

log_√5x = 2

Answer

Given,

⇒ log_√5x = 2

x = $(\sqrt{5})^2$ = 5.

Hence, x = 5.

Solve the following equation for x:

log_x 0.001 = -3

Answer

Given,

⇒ log_x 0.001 = -3

⇒ x^-3 = 0.001

⇒ x^-3 = $\dfrac{1}{1000} = \dfrac{1}{10^3}$

⇒ x^-3 = 10^-3

⇒ x = 10.

Hence, x = 10.

Solve the following equation for x:

log_√3(x + 1) = 2

Answer

Given,

⇒ log_√3(x + 1) = 2

⇒ (x + 1) = $(\sqrt{3})^2$

⇒ (x + 1) = 3

⇒ x = 2.

Hence, x = 2.

Solve the following equation for x:

log₄(2x + 3) = $\dfrac{3}{2}$

Answer

Given,

⇒ log₄(2x + 3) = $\dfrac{3}{2}$

⇒ 2x + 3 = $4^{\dfrac{3}{2}}$

⇒ 2x + 3 = $(2^2)^{\dfrac{3}{2}}$

⇒ 2x + 3 = 2³

⇒ 2x + 3 = 8

⇒ 2x = 5

⇒ x = $\dfrac{5}{2}$.

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

Solve the following equation for x:

$\text{log}_{\sqrt[3]{2}}$x = 3

Answer

Given,

$\Rightarrow \text{log}_{\sqrt[3]{2}}x = 3 \\[1em] \Rightarrow x = (\sqrt[3]{2})^3 \\[1em] \Rightarrow x = (2^{\dfrac{1}{3}})^3 \\[1em] \Rightarrow x = 2.$

Hence, x = 2.

Solve the following equation for x:

log₂(x² - 1) = 3

Answer

Given,

⇒ $\text{log}_2(x^2 - 1)$ = 3

⇒ x² - 1 = 2³

⇒ x² - 1 = 8

⇒ x² = 9

⇒ x = $\sqrt{9} = \pm3$.

Hence, x = ±3.

Solve the following equation for x:

log x = -1

Answer

Given,

⇒ log x = -1

⇒ x = 10^-1

⇒ x = $\dfrac{1}{10}.$

Hence, x = $\dfrac{1}{10}$.

Solve the following equation for x:

log(2x - 3) = 1

Answer

Given,

⇒ log(2x - 3) = 1

⇒ 2x - 3 = 10¹

⇒ 2x - 3 = 10

⇒ 2x = 13

⇒ x = $\dfrac{13}{2} = 6\dfrac{1}{2}.$

Hence, x = $6\dfrac{1}{2}.$

Solve the following equation for x:

log x = -2, 0, $\dfrac{1}{3}$.

Answer

Given,

⇒ log x = -2

⇒ x = 10^-2 = $\dfrac{1}{10^2} = \dfrac{1}{100}$.

⇒ log x = 0

⇒ x = 10⁰ = 1.

⇒ log x = $\dfrac{1}{3}$

⇒ x = $10^{\dfrac{1}{3}} = \sqrt[3]{10}.$

Hence, x = $\dfrac{1}{100}, 1, \sqrt[3]{10}.$

Given log₁₀a = b, express 10^{2b - 3} in terms of a.

Answer

Given,

⇒ log₁₀a = b

∴ a = 10^b.

Simplifying 10^{2b - 3} we get,

⇒ 10^{2b - 3} = 10^2b.10^-3

= $\dfrac{10^{2b}}{10^3} = \dfrac{(10^b)^2}{10^3}$

= $\dfrac{a^2}{10^3} = \dfrac{a^2}{1000}$.

Hence, 10^{2b - 3} = $\dfrac{a^2}{1000}.$

Given log₁₀x = a, log₁₀y = b and log₁₀z = c,

(i) write down 10^{2a - 3} in terms of x.

(ii) write down 10^{3b - 1} in terms of y.

(iii) if log₁₀P = 2a + $\dfrac{b}{2}$ - 3c, express P in terms of x, y and z.

Answer

(i) Given,

⇒ log₁₀x = a

∴ x = 10^a.

Simplifying 10^{2a - 3} we get,

⇒ 10^{2a - 3} = 10^2a.10^-3

= $\dfrac{10^{2a}}{10^3} = \dfrac{(10^a)^2}{10^3}$

= $\dfrac{x^2}{10^3} = \dfrac{x^2}{1000}$.

Hence, 10^{2a - 3} = $\dfrac{x^2}{1000}.$

(ii) Given,

⇒ log₁₀y = b

∴ y = 10^b.

Simplifying 10^{3b - 1} we get,

⇒ 10^{3b - 1} = 10^3b.10^-1

= $\dfrac{10^{3b}}{10^1} = \dfrac{(10^b)^3}{10}$

= $\dfrac{y^3}{10}$.

Hence, 10^{3b - 1} = $\dfrac{y^3}{10}.$

(iii) Given,

log₁₀z = c

⇒ z = 10^c.

log₁₀P = 2a + $\dfrac{b}{2}$ - 3c

$\Rightarrow P = 10^{2a + \frac{b}{2} - 3c} \\[1em] = 10^{2a}.10^{\frac{b}{2}}.10^{-3c} \\[1em] = (10^a)^2.(10^b)^{\frac{1}{2}}.(10^c)^{-3} \\[1em] = (x)^2.(y)^{\frac{1}{2}}.(z)^{-3} \\[1em] = \dfrac{x^2\sqrt{y}}{z^3}.$

Hence, P = $\dfrac{x^2\sqrt{y}}{z^3}.$

If log₁₀ x = a and log₁₀ y = b, find the value of xy.

Answer

Given,

log₁₀x = a

⇒ x = 10^a.

log₁₀y = b

⇒ y = 10^b

xy = 10^a.10^b = 10^{a + b}.

Hence, xy = 10^{a + b}.

Given log₁₀a = m and log₁₀b = n, express $\dfrac{a^3}{b^2}$ in terms of m and n.

Answer

Given,

log₁₀a = m

⇒ a = 10^m

log₁₀b = n

⇒ b = 10ⁿ

$\Rightarrow \dfrac{a^3}{b^2} = \dfrac{(10^m)^3}{(10^n)^2} \\[1em] = \dfrac{10^{3m}}{10^{2n}} = 10^{3m - 2n}.$

Hence, $\dfrac{a^3}{b^2}$ = 10^{3m - 2n}.

Given log₁₀x = 2a and log₁₀y = $\dfrac{b}{2}$,

(i) write 10^a in terms of x.

(ii) write 10^{2b + 1} in terms of y.

(iii) if log₁₀P = 3a - 3b, express P in terms of x and y.

Answer

(i) Given,

log₁₀x = 2a

⇒ x = 10^2a

⇒ x = (10^a)²

⇒ 10^a = $\sqrt{x}$.

Hence, 10^a = $\sqrt{x}$.

(ii) Given,

log₁₀y = $\dfrac{b}{2}$

$\Rightarrow y = 10^{\dfrac{b}{2}} \\[1em] \Rightarrow y = (10^b)^{\dfrac{1}{2}} \\[1em]$

Squaring both sides we get,

⇒ y² = 10^b

Simplifying 10^{2b + 1} we get,

10^{2b + 1} = 10^2b.10

= (10^b)².10

= (y²)².10

= 10y⁴.

Hence, 10^{2b + 1} = 10y⁴.

(iii) Given,

log₁₀P = 3a - 2b

⇒ P = 10^{3a - 2b}

⇒ P = 10^3a.10^-2b

= (10^a)³.(10^b)^-2

$= (\sqrt{x})^3 \times (y^2)^{-2} \\[1em] = (x^3)^{\dfrac{1}{2}} \times \dfrac{1}{(y^2)^2} \\[1em] = \dfrac{x^{\dfrac{3}{2}}}{y^4}$

Hence, P = $\dfrac{x^{\dfrac{3}{2}}}{y^4}$.

If log₂y = x and log₃z = x, find 72^x in terms of y and z.

Answer

Given,

log₂y = x and log₃z = x

⇒ y = 2^x and z = 3^x.

(72)^x = (2³.3²)^x

= (2)^3x.(3)^2x

= (2^x)³.(3^x)²

= (y)³.(z)²

Hence, (72)^x = y³.z².

If log₂x = a and log₅y = a, write 100^{2a - 1} in terms of x and y.

Answer

Given,

log₂x = a and log₅y = a

⇒ x = 2^a and y = 5^a

100^{2a - 1} = 100^2a.100^-1

= (100^a)².100^-1

= [(2².5²)^a]².100^-1

= [(2^2a).(5^2a)]².100^-1

= [(2^a)².(5^a)²]².(100)^-1

= [x².y²]².100^-1

= $\dfrac{x^4y^4}{100}$.

Hence, 100^{2a - 1} = $\dfrac{x^4y^4}{100}$.