solve check for extraneous roots 2(5^6x)-9(5^4x)+10(5^2x)-3=0, e log3(x)+log2(x)=5 and log(logx)=0

Let's solve each of the given equations and check for extraneous roots:

### Equation 1: \( 2(5^{6x}) - 9(5^{4x}) + 10(5^{2x}) - 3 = 0 \)

1. Let \( y = 5^{2x} \). Then the equation becomes:
   \[
   2y^3 - 9y^2 + 10y - 3 = 0
   \]

2. Solve the cubic equation for \( y \):
   \[
   2y^3 - 9y^2 + 10y - 3 = 0
   \]

   This is a cubic equation. Let's find its roots by factoring or using the Rational Root Theorem and synthetic division.

   After testing some potential rational roots, we find that \( y = 1 \) is a root.
   
3. Perform synthetic division to factor the cubic polynomial:

   \[
   \begin{array}{r|rrrr}
   1 & 2 & -9 & 10 & -3 \\
     &   &  2 & -7 &  3 \\
   \hline
     & 2 & -7 &  3 &  0 \\
   \end{array}
   \]

   The polynomial can be factored as:
   \[
   2y^3 - 9y^2 + 10y - 3 = (y - 1)(2y^2 - 7y + 3)
   \]

4. Solve the quadratic equation \( 2y^2 - 7y + 3 = 0 \) using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   y = \frac{7 \pm \sqrt{49 - 24}}{4} = \frac{7 \pm \sqrt{25}}{4} = \frac{7 \pm 5}{4}
   \]
   Therefore,
   \[
   y = 3 \quad \text{or} \quad y = \frac{1}{2}
   \]

   So, the roots are \( y = 1, y = 3, y = \frac{1}{2} \).

5. Convert back to \( x \):
   \[
   5^{2x} = 1 \implies 2x = 0 \implies x = 0
   \]
   \[
   5^{2x} = 3 \implies 2x = \log_5 3 \implies x = \frac{\log_5 3}{2}
   \]
   \[
   5^{2x} = \frac{1}{2} \implies 2x = \log_5 \left(\frac{1}{2}\right) \implies x = \frac{\log_5 \left(\frac{1}{2}\right)}{2}
   \]

### Equation 2: \( \log_3(x) + \log_2(x) = 5 \)

1. Use the change of base formula and combine the logarithms:
   \[
   \log_3(x) + \log_2(x) = 5
   \]
   \[
   \frac{\log(x)}{\log(3)} + \frac{\log(x)}{\log(2)} = 5
   \]
   \[
   \log(x) \left( \frac{1}{\log(3)} + \frac{1}{\log(2)} \right) = 5
   \]
   Let \( k = \log(x) \):
   \[
   k \left( \frac{1}{\log(3)} + \frac{1}{\log(2)} \right) = 5
   \]
   \[
   k \left( \frac{\log(2) + \log(3)}{\log(2)\log(3)} \right) = 5
   \]
   \[
   k \left( \frac{\log(6)}{\log(2)\log(3)} \right) = 5
   \]
   \[
   k = 5 \frac{\log(2)\log(3)}{\log(6)}
   \]
   \[
   k = 5 \frac{\log(6)}{\log(6)} = 5
   \]
   So, \( \log(x) = 5 \), which means:
   \[
   x = 10^5 = 100000
   \]

### Equation 3: \( \log(\log(x)) = 0 \)

1. Solve for \( x \):
   \[
   \log(\log(x)) = 0
   \]
   \[
   \log(x) = 10^0 = 1
   \]
   \[
   x = 10^1 = 10
   \]

### Summary of Solutions
1. \( 2(5^{6x}) - 9(5^{4x}) + 10(5^{2x}) - 3 = 0 \):
   - \( x = 0 \)
   - \( x = \frac{\log_5 3}{2} \)
   - \( x = \frac{\log_5 \left(\frac{1}{2}\right)}{2} \)

2. \( \log_3(x) + \log_2(x) = 5 \):
   - \( x = 100000 \)

3. \( \log(\log(x)) = 0 \):
   - \( x = 10 \)

Would you like details on any of these steps or have any questions?

### Relative Questions:
1. How do you derive the quadratic formula?
2. What is the Rational Root Theorem and how is it used?
3. Can you explain synthetic division in more detail?
4. How do you change the base of a logarithm?
5. What are the properties of logarithms used in solving logarithmic equations?
6. How do you determine if a root is extraneous?
7. What are the steps to solving a cubic equation?
8. How does the change of base formula for logarithms work?

### Tip:
When solving equations involving logarithms, always check the domain restrictions to ensure the solutions are valid within the domain of the original equation.

Explore solutions to complex equations involving exponentials and logarithms. Learn how to apply the quadratic formula, change of base formula for logarithms, and synthetic division to solve equations. Suitable for advanced high school students.

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