1. **Problem:** Solve for $x$ in the equation $\log_2(x) = 3$. 2. **Formula:** Recall that $\log_b(a) = c$ means $b^c = a$. 3. **Solution:** Using the formula, $\log_2(x) = 3$ implies $x = 2^3$. 4. **Evaluation:** Calculate $2^3 = 8$. 5. **Answer:** Therefore, $x = 8$. 1. **Problem:** Simplify $\log_5(25) + \log_5(4)$. 2. **Formula:** Use the logarithm property $\log_b(m) + \log_b(n) = \log_b(m \times n)$. 3. **Solution:** $\log_5(25) + \log_5(4) = \log_5(25 \times 4) = \log_5(100)$. 4. **Evaluation:** Since $25 = 5^2$, $\log_5(25) = 2$, and $\log_5(100)$ can be simplified as $\log_5(5^2 \times 4)$ but better to leave as is or approximate. 5. **Answer:** The simplified form is $\log_5(100)$. 1. **Problem:** Solve for $x$: $\log(x) + \log(x-3) = 1$ (log base 10). 2. **Formula:** Use $\log(a) + \log(b) = \log(ab)$. 3. **Solution:** $\log(x) + \log(x-3) = \log(x(x-3)) = 1$. 4. **Rewrite:** $\log(x^2 - 3x) = 1$ means $x^2 - 3x = 10^1 = 10$. 5. **Equation:** $x^2 - 3x - 10 = 0$. 6. **Solve quadratic:** Using the quadratic formula $x = \frac{3 \pm \sqrt{9 + 40}}{2} = \frac{3 \pm \sqrt{49}}{2} = \frac{3 \pm 7}{2}$. 7. **Solutions:** $x = 5$ or $x = -2$. 8. **Check domain:** $x$ must be greater than 0 and $x-3 > 0$, so $x > 3$. 9. **Valid solution:** $x = 5$. 1. **Problem:** Express $\log_3(81)$ in terms of a number. 2. **Recall:** $81 = 3^4$. 3. **Solution:** $\log_3(81) = \log_3(3^4) = 4$. 1. **Problem:** Solve for $x$: $\log_2(x) + \log_2(x-2) = 3$. 2. **Formula:** $\log_b(m) + \log_b(n) = \log_b(mn)$. 3. **Rewrite:** $\log_2(x(x-2)) = 3$. 4. **Rewrite exponential:** $x(x-2) = 2^3 = 8$. 5. **Equation:** $x^2 - 2x - 8 = 0$. 6. **Solve quadratic:** $x = \frac{2 \pm \sqrt{4 + 32}}{2} = \frac{2 \pm \sqrt{36}}{2} = \frac{2 \pm 6}{2}$. 7. **Solutions:** $x = 4$ or $x = -2$. 8. **Domain check:** $x > 0$ and $x-2 > 0$ so $x > 2$. 9. **Valid solution:** $x = 4$.