Given that :$$40^{0.11} = x, 40.^{0.1} = y$$ and $$x^{z} = y^{2},$$ then the value of z is close to:​

Correct option is B

Given:

​$$40^{0.11} = x, \quad 40^{0.1} = y, \quad x^z = y^2$$ 

Value of z = ? 

Solution:

We already have:

​$$x = 40^{0.11}, \quad y = 40^{0.1}$$ 

Now, substitute the expressions for x and y into the equation $$\quad x^z = y^2$$ 

$$(40^{0.11})^z = (40^{0.1})^2$$ 

Using the properties of exponents $$(a^m)^n = a^{m \cdot n}$$​ we can simplify both sides of the equation: 

​$$40^{0.11z} = 40^{0.2}$$ 

Since the bases are the same, we can set the exponents equal to each other:

​$$0.11z=0.2$$ 

Now, solve for z: 

​$$z = \frac{0.2}{0.11} \approx 1.818$$ 

z = 1.82 (approx) 

Thus, the value of z is approximately 1.82.