https://file.aiquickdraw.com/mathbot/file/1727300999370caxtushv.jpg

Let's solve these one by one.

### 4. Solve for \( x \):
\[
\log_3(5x) - \log_3(3x - 2) = -1
\]

Using the logarithmic identity \( \log_a(m) - \log_a(n) = \log_a\left(\frac{m}{n}\right) \), the equation becomes:
\[
\log_3\left(\frac{5x}{3x - 2}\right) = -1
\]
This means:
\[
\frac{5x}{3x - 2} = 3^{-1} = \frac{1}{3}
\]
Now, cross-multiply:
\[
5x = \frac{1}{3}(3x - 2)
\]
\[
15x = 3x - 2
\]
\[
15x - 3x = -2
\]
\[
12x = -2
\]
\[
x = -\frac{2}{12} = -\frac{1}{6}
\]

So, the solution is:
\[
x = -\frac{1}{6}
\]

---

### 5. Solve for \( x \):
\[
\log_2(x^5) + \log_x(32) = 10
\]

Using the logarithmic identity \( \log_a(b^n) = n \log_a(b) \), the equation becomes:
\[
5 \log_2(x) + \log_x(32) = 10
\]

Next, we know that \( \log_x(32) = \frac{1}{\log_2(x)} \), because \( 32 = 2^5 \), and by change of base formula:
\[
5 \log_2(x) + \frac{1}{\log_2(x)} = 10
\]
Let \( y = \log_2(x) \). The equation becomes:
\[
5y + \frac{1}{y} = 10
\]
Multiply both sides by \( y \) to get rid of the fraction:
\[
5y^2 + 1 = 10y
\]
Rearrange this quadratic:
\[
5y^2 - 10y + 1 = 0
\]
Solve this using the quadratic formula:
\[
y = \frac{-(-10) \pm \sqrt{(-10)^2 - 4(5)(1)}}{2(5)}
\]
\[
y = \frac{10 \pm \sqrt{100 - 20}}{10}
\]
\[
y = \frac{10 \pm \sqrt{80}}{10}
\]
\[
y = \frac{10 \pm 4\sqrt{5}}{10}
\]
\[
y = 1 \pm \frac{2\sqrt{5}}{5}
\]

Since \( y = \log_2(x) \), we can compute \( x \) using \( x = 2^y \).

---

### 6. Solve for \( x \):
\[
2(5^{x+1}) = 1 + \frac{3}{5^x}
\]

First, simplify:
\[
2 \cdot 5^{x+1} = 2 \cdot 5 \cdot 5^x = 10 \cdot 5^x
\]
The equation becomes:
\[
10 \cdot 5^x = 1 + \frac{3}{5^x}
\]
Multiply both sides by \( 5^x \) to clear the fraction:
\[
10 \cdot (5^x)^2 = 5^x + 3
\]
Let \( y = 5^x \). The equation becomes:
\[
10y^2 = y + 3
\]
Rearrange this into a quadratic equation:
\[
10y^2 - y - 3 = 0
\]
Solve using the quadratic formula:
\[
y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(10)(-3)}}{2(10)}
\]
\[
y = \frac{1 \pm \sqrt{1 + 120}}{20}
\]
\[
y = \frac{1 \pm \sqrt{121}}{20}
\]
\[
y = \frac{1 \pm 11}{20}
\]
So, \( y = \frac{12}{20} = \frac{3}{5} \) or \( y = \frac{-10}{20} = -\frac{1}{2} \).

Since \( y = 5^x \), we solve \( 5^x = \frac{3}{5} \), which gives:
\[
x = -\log_5\left(\frac{5}{3}\right)
\]

---

### 7. Find the exact value of \( x \) for the equation:
\[
(3^x)(4^{2x+1}) = 6^{x+2}
\]

First, express everything in terms of powers of 2 and 3:
\[
3^x \cdot (2^2)^{2x+1} = (2 \cdot 3)^{x+2}
\]
Simplify the exponents:
\[
3^x \cdot 2^{4x+2} = 2^{x+2} \cdot 3^{x+2}
\]
Now, divide both sides by \( 3^x \):
\[
2^{4x+2} = 2^{x+2} \cdot 3^2
\]
\[
2^{4x+2} = 2^{x+2} \cdot 9
\]
Divide both sides by \( 2^{x+2} \):
\[
2^{3x} = 9
\]
Take the logarithm base 2:
\[
3x = \log_2(9)
\]
\[
x = \frac{\log_2(9)}{3}
\]

Now express the answer in the form \( \frac{\ln a}{\ln b} \):
\[
x = \frac{\ln 9}{3 \ln 2}
\]
\[
x = \frac{\ln 3^2}{3 \ln 2} = \frac{2 \ln 3}{3 \ln 2}
\]

So the exact value is:
\[
x = \frac{\ln 3^2}{3 \ln 2}
\]

---

Let me know if you need further details on any of these solutions.

**Related Questions:**

1. Can we apply the properties of logarithms to simplify expressions involving multiple logarithms?
2. How do we solve quadratic equations involving logarithms?
3. What are the logarithmic identities and how can they simplify equations?
4. How does the quadratic formula help in solving transcendental equations?
5. What is the change of base formula in logarithms, and when is it useful?

**Tip:** Always check if an equation can be simplified using logarithmic identities before proceeding to more complex steps.

Solve the following math problems:
4. log_3(5x) - log_3(3x - 2) = -1
5. log_2(x^5) + log_x(32) = 10
6. 2(5^{x+1}) = 1 + 3/5^x
7. (3^x)(4^{2x+1}) = 6^{x+2}

Learn how to solve logarithmic and exponential equations involving quadratic forms. Detailed step-by-step solutions are provided for problems such as log_3(5x) - log_3(3x - 2) = -1 and (3^x)(4^{2x+1}) = 6^{x+2), using key concepts like the change of base formula, quadratic formula, and logarithmic properties.

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