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

(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.