lg x +lg(x-3)=1
To solve the equation \( \log x + \log(x - 3) = 1 \), we can use the properties of logarithms. Specifically, we can combine the logarithmic terms on the left-hand side using the product rule:
\[ \log x + \log(x - 3) = \log(x(x - 3)) \]
Substituting this back into the equation gives us:
\[ \log(x(x - 3)) = 1 \]
Now, we can convert the logarithmic equation to its exponential form:
\[ x(x - 3) = 10^1 \]
This simplifies to:
\[ x(x - 3) = 10 \]
Expanding the left side, we get:
\[ x^2 - 3x = 10 \]
Now, rearranging this into standard form gives us:
\[ x^2 - 3x - 10 = 0 \]
Next, we can solve this quadratic equation using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
where \( a = 1 \), \( b = -3 \), and \( c = -10 \).
Calculating the discriminant:
\[ b^2 - 4ac = (-3)^2 - 4(1)(-10) = 9 + 40 = 49 \]
Now substitute back into the quadratic formula:
\[ x = \frac{-(-3) \pm \sqrt{49}}{2(1)} = \frac{3 \pm 7}{2} \]
This gives us two possible solutions:
- \( x = \frac{3 + 7}{2} = \frac{10}{2} = 5 \)
- \( x = \frac{3 - 7}{2} = \frac{-4}{2} = -2 \)
Since the logarithm is only defined for positive arguments, we discard \( x = -2 \).
Thus, the only valid solution is:
\[ x = 5 \]
Lastly, we can verify the solution by substituting \( x = 5 \) back into the original equation:
\[ \log 5 + \log(5 - 3) = \log 5 + \log 2 = \log(5 \cdot 2) = \log 10 = 1 \]
Therefore, the solution satisfies the original equation. The final answer is:
\[ \boxed{5} \]
