We are given the equation:

$\log_7{x} + \log_7{(3x - 2)} = 0$

Step 1: Apply the Logarithmic Property

We can use the logarithmic property:

$\log_b{A} + \log_b{B} = \log_b{(A \cdot B)}$

Thus, the equation becomes:

$\log_7{(x \cdot (3x - 2))} = 0$

Step 2: Simplify the Equation

Now simplify the expression inside the logarithm:

$\log_7{(x(3x - 2))} = 0$

This simplifies further to:

$\log_7{(3x^2 - 2x)} = 0$

Step 3: Eliminate the Logarithm

Since $\log_7{A} = 0$ implies $A = 1$, we can write:

$3x^2 - 2x = 1$

Step 4: Rearrange the Equation

Rearrange the equation to form a quadratic:

$3x^2 - 2x - 1 = 0$

Step 5: Solve the Quadratic Equation

Now, solve the quadratic equation using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

For $3x^2 - 2x - 1 = 0$, the coefficients are $a = 3$, $b = -2$, and $c = -1$. Substituting these into the quadratic formula:

$x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(3)(-1)}}{2(3)}$

$x = \frac{2 \pm \sqrt{4 + 12}}{6}$

$x = \frac{2 \pm \sqrt{16}}{6}$

$x = \frac{2 \pm 4}{6}$

Step 6: Calculate the Solutions

Now solve for the two possible values of $x$:

$x = \frac{2 + 4}{6} = \frac{6}{6} = 1$

$x = \frac{2 - 4}{6} = \frac{-2}{6} = -\frac{1}{3}$

Step 7: Check for Valid Solutions

Since logarithms are only defined for positive values of their arguments, we need to check which solutions are valid:

• For $x = 1$, both $\log_7(1)$ and $\log_7(3(1) - 2)$ are valid.
• For $x = -\frac{1}{3}$, the logarithmic terms become undefined (since $x$ must be positive).

Thus, the only valid solution is:

$x = 1$

Final Answer:

$x = 1$

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

5 Related Questions:

1. How do you solve logarithmic equations with different bases?
2. Can you solve $\log_2{x} + \log_2{(x - 1)} = 3$?
3. What are the properties of logarithms used in solving equations?
4. How can you graph logarithmic functions and their transformations?
5. How do you solve logarithmic inequalities?

Tip:

Always check the domain of the logarithmic equation to ensure solutions make the argument of the logarithms positive.