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

(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.