This example explains how to solve the logarithmic equation:

$\log_5(x^2 + 1) = \log_5(x - 2) + \log_5(x + 3).$

Steps to Solve:

1. Combine the Right-Hand Side Using the Logarithmic Property:

   • Use the property: $\log_a(AB) = \log_a(A) + \log_a(B)$.
   • Combine $\log_5(x - 2) + \log_5(x + 3)$ into $\log_5[(x - 2)(x + 3)]$.

   The equation becomes: $\log_5(x^2 + 1) = \log_5[(x - 2)(x + 3)].$

2. Simplify Using the One-to-One Property:

   • Since $\log_5(A) = \log_5(B)$ implies $A = B$, you can remove the logarithms: $x^2 + 1 = (x - 2)(x + 3).$

3. Expand and Simplify the Right-Hand Side:

   • Expand $(x - 2)(x + 3)$: $x^2 + 1 = x^2 + x - 6.$

4. Solve for $x$:

   • Cancel $x^2$ from both sides: $1 = x - 6.$
   • Solve for $x$: $x = 7.$

5. Verify the Solution:

   • Substitute $x = 7$ back into the original equation to check if it satisfies the equation:
     • Left-hand side: $\log_5(7^2 + 1) = \log_5(49 + 1) = \log_5(50)$,
     • Right-hand side: $\log_5(7 - 2) + \log_5(7 + 3) = \log_5(5) + \log_5(10) = \log_5(50)$,
     • Both sides are equal, so $x = 7$ is valid.

Final Solution:

$x = 7.$

Would you like a more detailed breakdown of any specific step? Let me know!

5 Related Questions:

1. What are the key properties of logarithms used in solving equations like this one?
2. Can logarithms with different bases be solved similarly, or do special steps apply?
3. How do you handle logarithmic equations where there are extraneous solutions?
4. What happens if the argument inside a logarithm becomes negative or zero?
5. How does the one-to-one property of logarithms simplify logarithmic equations?

One Tip:

Always check your solution by substituting it back into the original equation to ensure it doesn't make any logarithmic arguments negative or undefined!