$$ f(x)=8 \log x-1 $$ Domain: Asymptote: $$ \square $$-intercept: $$ \square, \square $$-intercept: $$ (0, \square) $$

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**Domain:**  
Since the logarithm $$ \log x $$ is defined only for $$ x>0 $$, the domain is  
$$
(0, \infty)
$$

**Asymptote:**  
As $$ x\to 0^+ $$, the term $$ \log x $$ approaches $$-\infty$$ and therefore  
$$
f(x)=8\log x-1
$$
approaches $$-\infty$$. This indicates a vertical asymptote at  
$$
x=0.
$$
There is no horizontal asymptote because as $$ x\to\infty $$, $$ f(x)\to\infty $$.

**Intercepts:**

1. **x-intercept:**  
Set $$ f(x)=0 $$ and solve for $$ x $$:
$$
8\log x - 1=0 \quad\Longrightarrow\quad 8\log x = 1 \quad\Longrightarrow\quad \log x = \frac{1}{8}.
$$
Assuming the logarithm is the natural logarithm, we have
$$
x=e^{\frac{1}{8}}.
$$
Thus, the $$ x $$-intercept is
$$
\left(e^{\frac{1}{8}},\, 0\right).
$$

2. **y-intercept:**  
The $$ y $$-intercept is found by evaluating $$ f(0) $$. However, since $$ x=0 $$ is not in the domain, there is no $$ y $$-intercept.
**Domain:** All positive real numbers, $$ (0, \infty) $$. **Asymptote:** Vertical asymptote at $$ x=0 $$. **Intercepts:** - **x-intercept:** $$ \left(e^{\frac{1}{8}},\, 0\right) $$ - **y-intercept:** None

\( f(x)=8 \log x-1 \) Domain: Asymptote: \( \square \)-intercept: \( \square, \square \)-intercept: \( (0, \square) \)

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