If $\lim\limits_{x \to 0}\dfrac{45^x - 9^x - 5^x + 1}{(k^x - 1)(3^x - 1)} = 2$, then the value of $k$ is ...

The given limit is $\lim\limits_{x \to 0}\dfrac{45^x - 9^x - 5^x + 1}{(k^x - 1)(3^x - 1)} = 2$

Factorizing the numerator:

$45^x - 9^x - 5^x + 1 = 9^x(5^x - 1) - 1(5^x - 1) = (9^x - 1)(5^x - 1)$

Substituting this back into the limit:

$\lim\limits_{x \to 0}\dfrac{(9^x - 1)(5^x - 1)}{(k^x - 1)(3^x - 1)} = 2$

Dividing the numerator and the denominator by $x^2$:

$\lim\limits_{x \to 0}\dfrac{\dfrac{9^x - 1}{x} \cdot \dfrac{5^x - 1}{x}}{\dfrac{k^x - 1}{x} \cdot \dfrac{3^x - 1}{x}} = 2$

Using the standard limit $\lim\limits_{x \to 0}\dfrac{a^x - 1}{x} = \ln a$, we get:

$\dfrac{\ln 9 \cdot \ln 5}{\ln k \cdot \ln 3} = 2$

Since $\ln 9 = \ln(3^2) = 2 \ln 3$, the equation becomes:

$\dfrac{2 \ln 3 \cdot \ln 5}{\ln k \cdot \ln 3} = 2$

$\dfrac{2 \ln 5}{\ln k} = 2$

$\ln k = \ln 5$

$k = 5$

Answer: $5$