Logarithmic Equation B6Ef7C

1. **State the problem:** Solve the logarithmic equation $$2 \log x = \log (8x - 160) + 1$$ for $$x$$. 2. **Recall logarithm properties:** - $$a \log b = \log b^a$$ - $$\log a + \log b = \log (ab)$$ - The equation can be manipulated using these properties. 3. **Rewrite the equation:** $$2 \log x = \log (8x - 160) + 1$$ Rewrite $$2 \log x$$ as $$\log x^2$$: $$\log x^2 = \log (8x - 160) + 1$$ 4. **Express the constant 1 as a logarithm:** Since $$1 = \log 10$$ (assuming base 10 logarithm), rewrite: $$\log x^2 = \log (8x - 160) + \log 10$$ 5. **Combine the right side logarithms:** $$\log x^2 = \log [10(8x - 160)]$$ 6. **Set the arguments equal:** Since $$\log A = \log B$$ implies $$A = B$$, $$x^2 = 10(8x - 160)$$ 7. **Simplify the equation:** $$x^2 = 80x - 1600$$ Bring all terms to one side: $$x^2 - 80x + 1600 = 0$$ 8. **Factor the quadratic:** $$x^2 - 80x + 1600 = (x - 40)^2 = 0$$ 9. **Solve for $$x$$:** $$x - 40 = 0 \implies x = 40$$ 10. **Check the solution:** - Check the domain: $$x > 0$$ and $$8x - 160 > 0 \Rightarrow 8x > 160 \Rightarrow x > 20$$. - Since $$x = 40$$ satisfies both, it is valid. **Final answer:** $$\boxed{40}$$