To solve the equation $\ln(x^2 - 10) + \ln(9) = \ln(10)$, let's follow these steps using the one-to-one property of logarithms.

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Step 1: Combine the logarithms on the left-hand side

Using the logarithmic property $\ln(a) + \ln(b) = \ln(a \cdot b)$, we combine the left-hand side:

$\ln((x^2 - 10) \cdot 9) = \ln(10)$

This simplifies to:

$\ln(9(x^2 - 10)) = \ln(10)$

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Step 2: Use the one-to-one property of logarithms

The one-to-one property states that if $\ln(A) = \ln(B)$, then $A = B$. Apply this property:

$9(x^2 - 10) = 10$

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Step 3: Solve the quadratic equation

Expand and rearrange:

$9x^2 - 90 = 10$

$9x^2 - 100 = 0$

Factorize or solve directly:

$x^2 = \frac{100}{9}$

$x = \pm \frac{10}{3}$

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Step 4: Check for extraneous solutions

Logarithms are undefined for negative or zero arguments. For $\ln(x^2 - 10)$, the argument $x^2 - 10$ must be positive:

$x^2 - 10 > 0 \quad \implies \quad x^2 > 10$

This is satisfied by both $x = \pm \frac{10}{3}$ since $(\frac{10}{3})^2 = \frac{100}{9}$.

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Final Answer:

$x = \pm \frac{10}{3}$

You can write the solutions as:

$x = \frac{10}{3}; -\frac{10}{3}$

Would you like a detailed explanation of any step or concept? 😊

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Related Questions:

1. What is the one-to-one property of logarithms, and when is it used?
2. How do you solve quadratic equations like $9x^2 - 100 = 0$?
3. What happens if a logarithmic argument is negative or zero?
4. Can you solve $\ln(x+2) + \ln(x-1) = \ln(5)$?
5. How do properties of logarithms simplify solving equations?

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Tip:

Always check the domain of the logarithmic functions to ensure arguments are positive.