$$ f(x)=\log _{2}(x+3) $$ and $$ g(x)=\log _{2}(3 x+1) $$. (a) Solve $$ f(x)=4 $$. What point is on the graph of $$ f $$ ? (b) Solve $$ g(x)=4 $$. What point is on the graph of $$ g $$ ? (c) Solve $$ f(x)=g(x) $$. Do the graphs of $$ f $$ and $$ g $$ intersect? If so, where? (d) Solve $$ (f+g)(x)=7 $$. (e) Solve $$ (f-g)(x)=6 $$.

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Let's solve each part step by step.

---

### **(a) Solve $$ f(x) = 4 $$. What point is on the graph of $$ f $$?**

**Given:**
$$ f(x) = \log_{2}(x + 3) $$
$$ f(x) = 4 $$

**Solution:**
Set $$ f(x) = 4 $$:
$$
\log_{2}(x + 3) = 4
$$
Convert the logarithmic equation to its exponential form:
$$
x + 3 = 2^4
$$
$$
x + 3 = 16
$$
$$
x = 16 - 3
$$
$$
x = 13
$$

**Point on the graph:**
$$
(13, 4)
$$

---

### **(b) Solve $$ g(x) = 4 $$. What point is on the graph of $$ g $$?**

**Given:**
$$ g(x) = \log_{2}(3x + 1) $$
$$ g(x) = 4 $$

**Solution:**
Set $$ g(x) = 4 $$:
$$
\log_{2}(3x + 1) = 4
$$
Convert to exponential form:
$$
3x + 1 = 2^4
$$
$$
3x + 1 = 16
$$
$$
3x = 16 - 1
$$
$$
3x = 15
$$
$$
x = \frac{15}{3}
$$
$$
x = 5
$$

**Point on the graph:**
$$
(5, 4)
$$

---

### **(c) Solve $$ f(x) = g(x) $$. Do the graphs of $$ f $$ and $$ g $$ intersect? If so, where?**

**Given:**
$$
f(x) = \log_{2}(x + 3)
$$
$$
g(x) = \log_{2}(3x + 1)
$$

**Solution:**
Set $$ f(x) = g(x) $$:
$$
\log_{2}(x + 3) = \log_{2}(3x + 1)
$$
Since the logarithmic functions are equal, their arguments must be equal (assuming they are within the domain):
$$
x + 3 = 3x + 1
$$
$$
x - 3x = 1 - 3
$$
$$
-2x = -2
$$
$$
x = 1
$$
Now, find $$ y $$ using $$ f(x) $$ or $$ g(x) $$:
$$
f(1) = \log_{2}(1 + 3) = \log_{2}(4) = 2
$$

**Intersection Point:**
$$
(1, 2)
$$

**Conclusion:**
Yes, the graphs of $$ f $$ and $$ g $$ intersect at the point $$ (1, 2) $$.

---

### **(d) Solve $$ (f + g)(x) = 7 $$.**

**Given:**
$$
(f + g)(x) = f(x) + g(x) = \log_{2}(x + 3) + \log_{2}(3x + 1) = 7
$$

**Solution:**
Combine the logarithms using the product rule:
$$
\log_{2}\left( (x + 3)(3x + 1) \right) = 7
$$
Convert to exponential form:
$$
(x + 3)(3x + 1) = 2^7
$$
$$
(x + 3)(3x + 1) = 128
$$
Expand the left side:
$$
3x^2 + x + 9x + 3 = 128
$$
$$
3x^2 + 10x + 3 = 128
$$
Subtract 128 from both sides:
$$
3x^2 + 10x - 125 = 0
$$
Solve the quadratic equation using the quadratic formula:
$$
x = \frac{-10 \pm \sqrt{10^2 - 4 \cdot 3 \cdot (-125)}}{2 \cdot 3}
$$
$$
x = \frac{-10 \pm \sqrt{100 + 1500}}{6}
$$
$$
x = \frac{-10 \pm \sqrt{1600}}{6}
$$
$$
x = \frac{-10 \pm 40}{6}
$$
This gives two solutions:
$$
x = \frac{30}{6} = 5 \quad \text{and} \quad x = \frac{-50}{6} = -\frac{25}{3} \approx -8.333
$$

**Domain Consideration:**
- For $$ f(x) = \log_{2}(x + 3) $$, $$ x + 3 > 0 $$ ⇒ $$ x > -3 $$.
- For $$ g(x) = \log_{2}(3x + 1) $$, $$ 3x + 1 > 0 $$ ⇒ $$ x > -\frac{1}{3} $$.

**Valid Solution:**
$$
x = 5
$$
(The other solution $$ x = -\frac{25}{3} $$ is less than $$ -\frac{1}{3} $$ and thus not in the domain.)

---

### **(e) Solve $$ (f - g)(x) = 6 $$.**

**Given:**
$$
(f - g)(x) = f(x) - g(x) = \log_{2}(x + 3) - \log_{2}(3x + 1) = 6
$$

**Solution:**
Use the logarithmic quotient rule:
$$
\log_{2}\left( \frac{x + 3}{3x + 1} \right) = 6
$$
Convert to exponential form:
$$
\frac{x + 3}{3x + 1} = 2^6
$$
$$
\frac{x + 3}{3x + 1} = 64
$$
Multiply both sides by $$ 3x + 1 $$:
$$
x + 3 = 64(3x + 1)
$$
$$
x + 3 = 192x + 64
$$
Bring like terms together:
$$
x - 192x = 64 - 3
$$
$$
-191x = 61
$$
$$
x = -\frac{61}{191}
$$
$$
x \approx -0.319
$$

**Domain Consideration:**
- For $$ f(x) = \log_{2}(x + 3) $$, $$ x > -3 $$.
- For $$ g(x) = \log_{2}(3x + 1) $$, $$ x > -\frac{1}{3} $$.
  
Check $$ x = -\frac{61}{191} $$ (approximately -0.319):
$$
-0.319 > -0.333 \quad (\text{which is } -\frac{1}{3})
$$
Thus, it is within the domain.

**Solution:**
$$
x = -\frac{61}{191} \approx -0.319
$$

---

### **Summary of Solutions:**
- **(a)** $$ x = 13 $$. Point: $$ (13, 4) $$
- **(b)** $$ x = 5 $$. Point: $$ (5, 4) $$
- **(c)** The graphs intersect at $$ (1, 2) $$.
- **(d)** $$ x = 5 $$
- **(e)** $$ x = -\frac{61}{191} \approx -0.319 $$
**(a) $$ x = 13 $$. Point: $$ (13, 4) $$** **(b) $$ x = 5 $$. Point: $$ (5, 4) $$** **(c) The graphs intersect at $$ (1, 2) $$.** **(d) $$ x = 5 $$** **(e) $$ x = -\frac{61}{191} \approx -0.319 $$**

and .
(a) Solve . What point is on the graph of ?
(b) Solve . What point is on the graph of ?
© Solve . Do the graphs of and intersect? If so, where?
(d) Solve .
(e) Solve .

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and . (a) Solve . What point is on the graph of ? (b) Solve . What point is on the graph of ? © Solve . Do the graphs of and intersect? If so, where? (d) Solve . (e) Solve .