The equation provided is:

$5 \log_5(x) + \log_x(5) = 9$

Let's solve this step-by-step:

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Step 1: Simplify $\log_5(x)$

Using the logarithmic property $\log_b(a) = \frac{\log(a)}{\log(b)}$, we write: $\log_5(x) = \frac{\log(x)}{\log(5)}.$

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Step 2: Simplify $\log_x(5)$

Similarly, for $\log_x(5)$, we write: $\log_x(5) = \frac{\log(5)}{\log(x)}.$

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Step 3: Substitute these into the equation

Substituting these into the original equation: $5 \cdot \frac{\log(x)}{\log(5)} + \frac{\log(5)}{\log(x)} = 9.$

Let $y = \log(x)$, so the equation becomes: $5 \cdot \frac{y}{\log(5)} + \frac{\log(5)}{y} = 9.$

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Step 4: Eliminate the fraction

Multiply through by $y \cdot \log(5)$ to eliminate the denominators: $5y^2 + (\log(5))^2 = 9y \cdot \log(5).$

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Step 5: Rearrange into a standard quadratic form

Reorganize the terms: $5y^2 - 9y \cdot \log(5) + (\log(5))^2 = 0.$

This is a quadratic equation in $y$, where:

• $a = 5$,
• $b = -9 \cdot \log(5)$,
• $c = (\log(5))^2$.

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Step 6: Solve using the quadratic formula

The quadratic formula is: $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.$

Substituting $a$, $b$, and $c$: $y = \frac{9 \cdot \log(5) \pm \sqrt{(9 \cdot \log(5))^2 - 4 \cdot 5 \cdot (\log(5))^2}}{2 \cdot 5}.$

Simplify: $y = \frac{9 \cdot \log(5) \pm \sqrt{81 \cdot (\log(5))^2 - 20 \cdot (\log(5))^2}}{10}.$

$y = \frac{9 \cdot \log(5) \pm \sqrt{61 \cdot (\log(5))^2}}{10}.$

$y = \frac{9 \cdot \log(5) \pm \log(5) \cdot \sqrt{61}}{10}.$

Factor out $\log(5)$: $y = \log(5) \cdot \frac{9 \pm \sqrt{61}}{10}.$

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Step 7: Back-substitute for $y = \log(x)$

$\log(x) = \log(5) \cdot \frac{9 \pm \sqrt{61}}{10}.$

Exponentiate both sides to solve for $x$: $x = 5^{\frac{9 \pm \sqrt{61}}{10}}.$

Thus, there are two solutions: $x = 5^{\frac{9 + \sqrt{61}}{10}} \quad \text{and} \quad x = 5^{\frac{9 - \sqrt{61}}{10}}.$

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Final Answer:

The solutions are: $x = 5^{\frac{9 + \sqrt{61}}{10}}, \quad x = 5^{\frac{9 - \sqrt{61}}{10}}.$

Would you like a deeper explanation of any step?

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Here are 5 related questions to explore:

1. How can the quadratic formula be used in logarithmic equations more generally?
2. What are the properties of logarithms and how are they applied in problem-solving?
3. How do changes of base for logarithms affect their use in equations?
4. What happens if $\log_x(5)$ is written directly as its inverse $(1/\log_5(x))$?
5. Can these solutions for $x$ be expressed in terms of decimals instead of logarithmic expressions?

Tip: Always check if logarithmic solutions satisfy the domain of the original equation (e.g., $x > 0, x \neq 1$).