1. **State the problem:** Solve the equation $$\log \sqrt{3x + 1} = 1 + \log \sqrt{2x - 3}$$ for $x$. 2. **Recall logarithm properties:** - $\log a^b = b \log a$ - $\log a - \log b = \log \frac{a}{b}$ - $\log 10 = 1$ (assuming base 10 logarithm) 3. **Rewrite the equation:** $$\log \sqrt{3x + 1} = 1 + \log \sqrt{2x - 3}$$ can be written as $$\log (3x + 1)^{\frac{1}{2}} = \log 10 + \log (2x - 3)^{\frac{1}{2}}$$ 4. **Use log addition rule:** $$\log (3x + 1)^{\frac{1}{2}} = \log \left(10 \times (2x - 3)^{\frac{1}{2}}\right)$$ 5. **Since $\log A = \log B$, then $A = B$: ** $$ (3x + 1)^{\frac{1}{2}} = 10 (2x - 3)^{\frac{1}{2}} $$ 6. **Square both sides to eliminate square roots:** $$ 3x + 1 = 100 (2x - 3) $$ 7. **Expand and simplify:** $$ 3x + 1 = 200x - 300 $$ 8. **Bring all terms to one side:** $$ 3x + 1 - 200x + 300 = 0 $$ $$ -197x + 301 = 0 $$ 9. **Solve for $x$:** $$ -197x = -301 $$ $$ x = \frac{301}{197} $$ 10. **Check domain restrictions:** - Inside the logarithms, arguments must be positive: $$ 3x + 1 > 0 \Rightarrow x > -\frac{1}{3} $$ $$ 2x - 3 > 0 \Rightarrow x > \frac{3}{2} $$ Since $x = \frac{301}{197} \approx 1.527 > 1.5$, it satisfies the domain. **Final answer:** $$ x = \frac{301}{197} $$