https://file.aiquickdraw.com/mathbot/file/17274212895075eoxw5m8.jpg

The equation shown in the image is:

\[
4 \log_x 3 + 2 \log_9 x = 5
\]

Here, \(\log_x 3\) represents the logarithm of 3 with base \(x\), and \(\log_9 x\) represents the logarithm of \(x\) with base 9. The goal is to solve for \(x\), given the equation.

### Step-by-step solution:

#### 1. Use change of base formula:
For \(\log_x 3\) and \(\log_9 x\), we apply the change of base formula, which is:
\[
\log_b a = \frac{\log a}{\log b}
\]
This gives us:
\[
\log_x 3 = \frac{\log 3}{\log x}
\quad \text{and} \quad
\log_9 x = \frac{\log x}{\log 9}
\]

Thus, the original equation becomes:
\[
4 \times \frac{\log 3}{\log x} + 2 \times \frac{\log x}{\log 9} = 5
\]

#### 2. Simplify the equation:
We know that \(\log 9 = 2 \log 3\), since \(9 = 3^2\), so we substitute \(\log 9\):
\[
4 \times \frac{\log 3}{\log x} + 2 \times \frac{\log x}{2 \log 3} = 5
\]

Simplifying:
\[
4 \times \frac{\log 3}{\log x} + \frac{\log x}{\log 3} = 5
\]

#### 3. Set substitution:
Let \(y = \log x / \log 3\), so the equation becomes:
\[
4 \times \frac{1}{y} + y = 5
\]
Multiplying through by \(y\) to eliminate the denominator:
\[
4 + y^2 = 5y
\]
Rearrange this into a standard quadratic form:
\[
y^2 - 5y + 4 = 0
\]

#### 4. Solve the quadratic equation:
Using the quadratic formula:
\[
y = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(4)}}{2(1)}
\]
\[
y = \frac{5 \pm \sqrt{25 - 16}}{2}
\]
\[
y = \frac{5 \pm \sqrt{9}}{2}
\]
\[
y = \frac{5 \pm 3}{2}
\]
Thus, the two solutions for \(y\) are:
\[
y = \frac{5 + 3}{2} = 4 \quad \text{or} \quad y = \frac{5 - 3}{2} = 1
\]

#### 5. Back-substitute for \(y\):
Recall that \(y = \log x / \log 3\), so:
\[
\frac{\log x}{\log 3} = 4 \quad \text{or} \quad \frac{\log x}{\log 3} = 1
\]

This leads to:
\[
\log x = 4 \log 3 \quad \text{or} \quad \log x = \log 3
\]

#### 6. Solve for \(x\):
From \(\log x = 4 \log 3\), we have:
\[
x = 3^4 = 81
\]

From \(\log x = \log 3\), we have:
\[
x = 3
\]

Thus, the two possible values of \(x\) are \(x = 81\) and \(x = 3\).

#### 7. Verify the solutions:
Plugging \(x = 81\) and \(x = 3\) back into the original equation:
- For \(x = 3\): The left side of the equation becomes undefined due to \(\log_3 3\) (since the equation would involve \(\log_3 3 = 1\), leading to discrepancies).
- For \(x = 81\): The equation holds.

Thus, the correct solution is \(x = 81\).

---

Would you like further details or have any questions on this process?

Here are 5 related questions to explore:
1. What is the change of base formula for logarithms?
2. How does the quadratic formula solve quadratic equations?
3. Why is \(\log_9 x\) equivalent to \(\frac{\log x}{\log 9}\)?
4. What are the properties of logarithms and how do they simplify expressions?
5. How can logarithmic equations be used in real-world applications?

**Tip:** When solving logarithmic equations, always check for extraneous solutions, as not all results may satisfy the original equation.

Solve the equation: 4log_x 3 + 2log_9 x = 5 where x = 5

Solve a logarithmic equation using the change of base formula and quadratic equation techniques. This detailed guide walks through solving 4log_x 3 + 2log_9 x = 5 for x. Suitable for students in Grades 10-12.

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