1. **State the problem:** Solve the logarithmic equation $$\log(x^2 + 5) - \log 3 = 1$$. 2. **Recall the logarithm property:** The difference of logarithms is the logarithm of a quotient: $$\log a - \log b = \log \left(\frac{a}{b}\right)$$. 3. **Apply the property:** $$\log \left(\frac{x^2 + 5}{3}\right) = 1$$. 4. **Rewrite the logarithmic equation in exponential form:** Since the base of the logarithm is 10 (common log), $$\frac{x^2 + 5}{3} = 10^1 = 10$$. 5. **Solve for $x^2$:** Multiply both sides by 3: $$x^2 + 5 = 30$$ Subtract 5: $$x^2 = 25$$. 6. **Find $x$:** Take the square root of both sides: $$x = \pm 5$$. 7. **Check for extraneous solutions:** Substitute $x = 5$ into the original logarithmic expressions: $$x^2 + 5 = 25 + 5 = 30 > 0$$ (valid) Substitute $x = -5$: $$(-5)^2 + 5 = 25 + 5 = 30 > 0$$ (valid) Both solutions are valid. **Final answer:** $$x = 5 \text{ or } x = -5$$