1. We are asked to solve for $x$ in logarithmic equations and round answers to 2 decimal places. 2. Recall the definition of logarithm: $\log_a b = c$ means $a^c = b$. 3. For part (a) $\log_2 x = 2$: Using the definition, $x = 2^2 = 4$. 4. For part (b) $\log_x 16 = 4$: Using the definition, $x^4 = 16$. Since $16 = 2^4$, we have $x^4 = 2^4$. Taking the fourth root, $x = \sqrt[4]{16} = 2$. 5. For part (c) $\log_x 0.1 = -1$: Using the definition, $x^{-1} = 0.1$. This means $\frac{1}{x} = 0.1$. Multiply both sides by $x$ and divide by $0.1$: $$\cancel{x} \times \frac{1}{\cancel{x}} = 0.1 \times x \Rightarrow 1 = 0.1 x$$ $$\Rightarrow x = \frac{1}{0.1} = 10$$. 6. For part (d) $\log_x 0.01 = -2$: Using the definition, $x^{-2} = 0.01$. This means $\frac{1}{x^2} = 0.01$. Multiply both sides by $x^2$ and divide by $0.01$: $$\cancel{x^2} \times \frac{1}{\cancel{x^2}} = 0.01 \times x^2 \Rightarrow 1 = 0.01 x^2$$ $$\Rightarrow x^2 = \frac{1}{0.01} = 100$$ Taking square root, $x = \sqrt{100} = 10$. 7. For part (e) $\log_4 x = 1$: Using the definition, $4^1 = x$. So, $x = 4$. 8. For part (f) $\log x = 0.23$ (assuming base 10): Using the definition, $10^{0.23} = x$. Calculate $x = 10^{0.23} \approx 1.70$ (rounded to 2 decimals). Final answers: (a) $x=4$ (b) $x=2$ (c) $x=10$ (d) $x=10$ (e) $x=4$ (f) $x \approx 1.70$