log√6(х-1)+log√6(x+4)=log√6 6

To solve the equation \( \log_{\sqrt{6}}(x - 1) + \log_{\sqrt{6}}(x + 4) = \log_{\sqrt{6}}(6) \), we can use the properties of logarithms. Specifically, we will use the property that states \( \log_b(a) + \log_b(c) = \log_b(ac) \).

Step 1: Combine the logarithms

Using the property of logarithms, we can rewrite the left side of the equation:

\[ \log_{\sqrt{6}}((x - 1)(x + 4)) = \log_{\sqrt{6}}(6) \]

Step 2: Exponentiate to remove the logarithm

We can now exponentiate both sides to remove the logarithm, giving us:

\[ (x - 1)(x + 4) = 6 \]

Step 3: Expand the left side

Next, we expand the left side:

\[ x^2 + 4x - x - 4 = 6 \]

This simplifies to:

\[ x^2 + 3x - 4 = 6 \]

Step 4: Move all terms to one side

Now we move 6 to the left side to set the equation to zero:

\[ x^2 + 3x - 10 = 0 \]

Step 5: Factor the quadratic equation

Next, we factor the quadratic equation:

\[ (x + 5)(x - 2) = 0 \]

Step 6: Solve for \( x \)

Setting each factor equal to zero gives us the solutions:

1. \( x + 5 = 0 \) ⟹ \( x = -5 \)
2. \( x - 2 = 0 \) ⟹ \( x = 2 \)

Step 7: Check for valid solutions

We need to check whether these solutions are valid in the context of the original logarithmic equations, which have restrictions (the arguments must be positive).

1. For \( x = -5 \):

   • \( x - 1 = -6 \) (invalid because it's negative)
   • \( x + 4 = -1 \) (invalid because it's negative)

2. For \( x = 2 \):

   • \( x - 1 = 1 \) (valid)
   • \( x + 4 = 6 \) (valid)

Since only \( x = 2 \) gives positive arguments in the logarithm, we include this as the solution.

Final Answer:

\[ \boxed{2} \]