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

(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.