If `f(x)={{:(((2^(x)-1)^(2)tan3x)/(xsin^(2)x)":",0ltxltpi//6),(lambda":",x=0):}` is continuous at x = 0, then the value of `(10sqrt(3lambda))/(ln2)` is equal to

Answer

To solve the problem step by step, we need to ensure that the function \( f(x) \) is continuous at \( x = 0 \). This means that the limit of \( f(x) \) as \( x \) approaches 0 must equal \( f(0) \), which is given as \( \lambda \).

### Step 1: Set up the limit
We start with the function defined for \( 0 < x < \frac{\pi}{6} \):
\[ f(x) = \frac{(2^x - 1)^2 \tan(3x)}{x \sin^2(x)} \]
We need to find:
\[ \lim_{x \to 0} f(x) = \lambda \]

### Step 2: Apply the limit
We can rewrite the limit:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{(2^x - 1)^2 \tan(3x)}{x \sin^2(x)} \]

### Step 3: Use known limits
We will use the following known limits:
1. \( \lim_{x \to 0} \frac{2^x - 1}{x} = \ln(2) \)
2. \( \lim_{x \to 0} \frac{\tan(3x)}{3x} = 1 \)
3. \( \lim_{x \to 0} \frac{\sin(x)}{x} = 1 \)

### Step 4: Rewrite the limit expression
We can manipulate the limit to use these known limits:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{(2^x - 1)^2}{x^2} \cdot \frac{\tan(3x)}{3x} \cdot \frac{3x}{\sin^2(x)} \]

### Step 5: Substitute the limits
Now substituting the limits:
1. \( (2^x - 1)^2 \) becomes \( (\ln(2) \cdot x)^2 = (\ln(2))^2 x^2 \)
2. \( \tan(3x) \) becomes \( 3x \) as \( x \to 0 \)
3. \( \sin^2(x) \) becomes \( x^2 \)

Thus, we can write:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{(\ln(2))^2 x^2 \cdot 3x}{x \cdot x^2} = \lim_{x \to 0} \frac{3(\ln(2))^2 x^2}{x^3} = 3(\ln(2))^2 \]

### Step 6: Set the limit equal to \( \lambda \)
Since we have established that:
\[ \lim_{x \to 0} f(x) = 3(\ln(2))^2 \]
we can set this equal to \( \lambda \):
\[ \lambda = 3(\ln(2))^2 \]

### Step 7: Find the value of \( \frac{10\sqrt{3\lambda}}{\ln(2)} \)
Now we substitute \( \lambda \) into the expression:
\[ \frac{10\sqrt{3\lambda}}{\ln(2)} = \frac{10\sqrt{3 \cdot 3(\ln(2))^2}}{\ln(2)} = \frac{10\sqrt{9(\ln(2))^2}}{\ln(2)} = \frac{10 \cdot 3\ln(2)}{\ln(2)} = 30 \]

### Final Answer
Thus, the final value is:
\[ \boxed{30} \]