Solve the given equation. \[ \log (4 x-3)+\log (x+2)=1 \] \[ x \approx \] (Type an integer or a decimal. Round to two decimal places as needed. Use a comma to separate answers as needed.)

Step 1 ：Given the equation \(\log (4 x-3)+\log (x+2)=1\)

Step 2 ：Combine the two logarithms into one using the property of logarithms that states that the sum of the logarithms of two numbers is equal to the logarithm of the product of those numbers. This gives us \(\log ((4x - 3)(x + 2)) = 1\)

Step 3 ：Rewrite the equation in exponential form using the property of logarithms that states that if \(\log_b(a) = c\), then \(b^c = a\). This gives us \((4x - 3)(x + 2) = 10\)

Step 4 ：Expand the left side of the equation to get a quadratic equation: \(4x^2 + 8x - 3x - 6 = 10\)

Step 5 ：Simplify the equation to get \(4x^2 + 5x - 16 = 0\)

Step 6 ：Solve the quadratic equation for x. The solution is approximately \(x = 0.98\)

Step 7 ：\(\boxed{x = 0.98}\)