To solve the given system of equations and expression, we will approach each one individually.

Equation 1

$\sqrt{x-2} = 2\sqrt{y+1}$

Square both sides to eliminate the square roots: $(\sqrt{x-2})^2 = (2\sqrt{y+1})^2$ $x - 2 = 4(y + 1)$ $x - 2 = 4y + 4$ $x = 4y + 6$ This is the relationship between $x$ and $y$.

Equation 2

$3^{(z-4)} + 1 = xy$

Equation 3

$\log(a) - 1 = \log_{y}(a - 16)$

Convert the logarithmic equation: $\log(a) - 1 = \log_{y}(a - 16)$ $\log(a) - \log(10) = \log_{y}(a - 16)$ $\log\left(\frac{a}{10}\right) = \log_{y}(a - 16)$

Using the change of base formula: $\frac{\log(a - 16)}{\log(y)} = \log\left(\frac{a}{10}\right)$

Integral and LCM Expression

$\int_{xx}^{a} y \, dt - \text{LCM}(xyza)$

The integral: $\int_{xx}^{a} y \, dt = y \left[t\right]_{xx}^{a}$ $y(a - xx)$

The expression becomes: $y(a - xx) - \text{LCM}(xyza)$

Summary

1. Equation 1 yields the relationship $x = 4y + 6$.
2. Equation 2: $3^{(z-4)} + 1 = xy$.
3. Equation 3: Relates $a$ and $y$ through logarithms.
4. Integral and LCM: $y(a - xx) - \text{LCM}(xyza)$.

Questions for Further Exploration:

1. What are the possible values for $x$ and $y$ from the first equation?
2. How can we solve Equation 2 for $z$ given values of $x$ and $y$?
3. How does changing the base affect the logarithmic relationship in Equation 3?
4. How do we calculate the least common multiple (LCM) of $xyza$?
5. How do different values of $a$ affect the integral result?

Tip:

Always check the domain of your variables to ensure the solutions are valid within the context of the original equations.