1. **State the problem:** Solve the logarithmic equation $$\log_{x+1}(7x - 3) = 2$$ for $x$. 2. **Recall the logarithm definition:** If $$\log_a b = c$$, then $$a^c = b$$. 3. **Apply the definition:** Here, $$a = x+1$$, $$b = 7x - 3$$, and $$c = 2$$, so $$$(x+1)^2 = 7x - 3$$ 4. **Expand the left side:** $$$(x+1)^2 = x^2 + 2x + 1$$ 5. **Rewrite the equation:** $$$x^2 + 2x + 1 = 7x - 3$$ 6. **Bring all terms to one side:** $$$x^2 + 2x + 1 - 7x + 3 = 0$$ $$$x^2 - 5x + 4 = 0$$ 7. **Factor the quadratic:** $$$x^2 - 5x + 4 = (x - 4)(x - 1) = 0$$ 8. **Solve for $x$:** $$$x - 4 = 0 \Rightarrow x = 4$$ $$$x - 1 = 0 \Rightarrow x = 1$$ 9. **Check domain restrictions:** - Base of logarithm $x+1 > 0$ and $x+1 \neq 1$ (base cannot be 1) - Argument $7x - 3 > 0$ For $x=4$: - Base: $4 + 1 = 5 > 0$ and $5 \neq 1$ (valid) - Argument: $7(4) - 3 = 28 - 3 = 25 > 0$ (valid) For $x=1$: - Base: $1 + 1 = 2 > 0$ and $2 \neq 1$ (valid) - Argument: $7(1) - 3 = 7 - 3 = 4 > 0$ (valid) 10. **Verify solutions by substitution:** - For $x=4$, $$\log_5 25 = 2$$ since $$5^2 = 25$$ (true) - For $x=1$, $$\log_2 4 = 2$$ since $$2^2 = 4$$ (true) **Final answer:** $$x = 1 \text{ or } x = 4$$