To solve the equation [tex]\(\log_3 x + \log_3 (x^2 + 2) = 1 + 2 \log_3 x\)[/tex], let's go through the solution step-by-step.
1. Combine the logarithmic terms on the left-hand side:
[tex]\[
\log_3 x + \log_3 (x^2 + 2) = \log_3 \left( x \cdot (x^2 + 2) \right)
\][/tex]
Thus, the equation becomes:
[tex]\[
\log_3 \left( x(x^2 + 2) \right) = 1 + 2 \log_3 x
\][/tex]
2. Simplify the expression inside the logarithm:
[tex]\[
x(x^2 + 2) = x^3 + 2x
\][/tex]
So the equation now is:
[tex]\[
\log_3 (x^3 + 2x) = 1 + 2 \log_3 x
\][/tex]
3. Express [tex]\(2 \log_3 x\)[/tex] as a single logarithm:
[tex]\[
2 \log_3 x = \log_3 x^2
\][/tex]
So the equation can be rewritten as:
[tex]\[
\log_3 (x^3 + 2x) = 1 + \log_3 x^2
\][/tex]
4. Use the property of logarithms to combine [tex]\(1\)[/tex] and [tex]\(\log_3 x^2\)[/tex]:
[tex]\[
1 + \log_3 x^2 = \log_3 3 + \log_3 x^2 = \log_3 (3x^2)
\][/tex]
Hence, the equation becomes:
[tex]\[
\log_3 (x^3 + 2x) = \log_3 (3x^2)
\][/tex]
5. Since the logarithmic functions with the same base are equal, their arguments must be equal:
[tex]\[
x^3 + 2x = 3x^2
\][/tex]
6. Solve the algebraic equation:
Rearrange the equation:
[tex]\[
x^3 + 2x - 3x^2 = 0
\][/tex]
Simplifying further:
[tex]\[
x^3 - 3x^2 + 2x = 0
\][/tex]
7. Factor out the common term [tex]\(x\)[/tex]:
[tex]\[
x(x^2 - 3x + 2) = 0
\][/tex]
This gives us:
[tex]\[
x = 0 \quad \text{or} \quad x^2 - 3x + 2 = 0
\][/tex]
8. Solve the quadratic equation:
[tex]\[
x^2 - 3x + 2 = 0
\][/tex]
Factorizing the quadratic equation:
[tex]\[
(x - 1)(x - 2) = 0
\][/tex]
This gives us:
[tex]\[
x = 1 \quad \text{or} \quad x = 2
\][/tex]
So, the solutions to the equation [tex]\(\log_3 x + \log_3 (x^2 + 2) = 1 + 2 \log_3 x\)[/tex] are:
[tex]\[
x = 1 \quad \text{and} \quad x = 2
\][/tex]
Negative values [tex]\(x = -2\)[/tex] and [tex]\(x = -1\)[/tex] are not valid solutions as logarithms of non-positive numbers are undefined in real numbers. Hence, the valid solutions are [tex]\(x = 1\)[/tex] and [tex]\(x = 2\)[/tex].
