To find the x-intercept of the function $g(x) = \log(x + 4)$, we need to determine the value of $x$ where $g(x) = 0$.

Step-by-Step Process:

1. Set $g(x) = 0$: $g(x) = \log(x + 4)$ For the x-intercept, we set $g(x) = 0$, so: $\log(x + 4) = 0$

2. Use the definition of logarithm: The logarithmic function $\log_b(y) = x$ means that $b^x = y$. In this case, we're working with the common logarithm (base 10), so we can rewrite the equation as: $10^0 = x + 4$ Since $10^0 = 1$, this simplifies to: $1 = x + 4$

3. Solve for $x$: $x = 1 - 4$ $x = -3$

Conclusion:

The x-intercept of $g(x) = \log(x + 4)$ is $x = -3$. This is the point where the graph of the function crosses the x-axis.

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Would you like a more detailed explanation of logarithms or how to work with transformations? Here are some related questions you might find interesting:

1. How does the graph of $g(x) = \log(x + 4)$ differ from the graph of $f(x) = \log(x)$?
2. How would you find the x-intercept of a logarithmic function with a different base, say base 2 or base $e$?
3. What is the domain of the function $g(x) = \log(x + 4)$?
4. How would you graph the function $g(x) = \log(x + 4)$?
5. What is the significance of the vertical asymptote in logarithmic functions?

Tip: When working with logarithmic equations, remember that logarithms are only defined for positive arguments, so always ensure that the expression inside the logarithm is greater than zero.