To find the $x$- and $y$-intercepts of the function $g(x) = 2\log\left(5(x-1)\right) + 4$, we will follow these steps:

1. Find the $y$-Intercept

The $y$-intercept occurs where $x = 0$.

Substitute $x = 0$ into the function: $g(0) = 2\log\left(5(0-1)\right) + 4 = 2\log(-5) + 4$

However, the logarithm of a negative number is undefined in the real number system. Therefore, there is no $y$-intercept for this function.

2. Find the $x$-Intercept

The $x$-intercept occurs where $g(x) = 0$.

Set the function equal to zero: $2\log\left(5(x-1)\right) + 4 = 0$

Subtract 4 from both sides: $2\log\left(5(x-1)\right) = -4$

Divide both sides by 2: $\log\left(5(x-1)\right) = -2$

Rewrite the equation in exponential form: $5(x-1) = 10^{-2}$

Simplify $10^{-2}$ to get: $5(x-1) = \frac{1}{100}$

Divide both sides by 5: $x-1 = \frac{1}{500}$

Add 1 to both sides: $x = 1 + \frac{1}{500} = \frac{500}{500} + \frac{1}{500} = \frac{501}{500}$

Final Answers:

• $y$-intercept: None (because the logarithm of a negative number is undefined).
• $x$-intercept: $x = \frac{501}{500}$ or approximately $x = 1.002$.

Would you like more details on any step or have any other questions?

Here are some related questions you might find interesting:

1. What are the domain and range of $g(x) = 2\log\left(5(x-1)\right) + 4$?
2. How would the function change if the base of the logarithm were different?
3. Can we find asymptotes for the function $g(x)$?
4. What are the critical points of $g(x) = 2\log\left(5(x-1)\right) + 4$?
5. How would you graph this function?
6. What happens to the intercepts if we add a constant to the function?
7. How does the function behave as $x$ approaches infinity?
8. What are the transformations applied to the base function $\log(x)$ in this equation?

Tip: When solving logarithmic equations, always check the domain to avoid taking the logarithm of a negative number or zero.