- Identify the domain condition \(x > \frac{1}{6}\) for the logarithms to be defined.
- Combine the logarithms into a single expression \(\log_7(6x^2 - x) = 1\).
- Solve the resulting quadratic equation \(6x^2 - x - 7 = 0\) to find potential roots \(x = \frac{7}{6}\) and \(x = -1\).
- Filter the results based on the domain constraint to arrive at the final solution \(\boxed{\frac{7}{6}}\).
Explanation
- Defining the Domain
We are given the logarithmic equation \(\log_7 x + \log_7(6x - 1) = 1\). Before solving, we must identify the domain of validity for the logarithmic expressions. For a logarithm \(\log_b(y)\) to be defined, the argument \(y\) must be positive.
- For \(\log_7 x\), we require \(x > 0\).
- For \(\log_7(6x - 1)\), we require \(6x - 1 > 0\), which simplifies to \(x > \frac{1}{6}\).
Combining these two conditions, the valid domain for \(x\) is \(x > \frac{1}{6}\). Any solution we find must satisfy this condition.
- Simplifying and Transforming the Equation
We use the logarithmic product rule, which states that \(\log_b A + \log_b B = \log_b(AB)\). Applying this to our equation, we combine the logarithms on the left side:
\(\log_7(x(6x - 1)) = 1\)
This gives us \(\log_7(6x^2 - x) = 1\).
By the definition of a logarithm, \(\log_b Y = C\) is equivalent to \(Y = b^C\). Applying this to our equation where \(b = 7\), \(Y = 6x^2 - x\), and \(C = 1\), we obtain:
\(6x^2 - x = 7^1\)
\(6x^2 - x = 7\)
Rearranging this into standard quadratic form \(ax^2 + bx + c = 0\):
\(6x^2 - x - 7 = 0\)
- Solving the Quadratic Equation
We need to solve the quadratic equation \(6x^2 - x - 7 = 0\). We can attempt to factor this expression:
We look for two numbers that multiply to \(6 \times (-7) = -42\) and add to \(-1\). These numbers are \(-7\) and \(6\).
Rewriting the middle term:
\(6x^2 + 6x - 7x - 7 = 0\)
Grouping the terms:
\(6x(x + 1) - 7(x + 1) = 0\)
\((6x - 7)(x + 1) = 0\)
This gives two potential solutions for \(x\):
- \(6x - 7 = 0 \implies x = \frac{7}{6}\)
- \(x + 1 = 0 \implies x = -1\)
- Verifying Solutions and Concluding
Now we check our potential solutions against the domain constraint established in Step 1, which is \(x > \frac{1}{6}\).
- For \(x = \frac{7}{6}\), we see that \(\frac{7}{6} \approx 1.167\), which is greater than \(\frac{1}{6} \approx 0.167\). This solution is valid.
- For \(x = -1\), this value does not satisfy the condition \(x > \frac{1}{6}\). Therefore, we must reject \(x = -1\).
The only valid solution is \(x = \frac{7}{6}\). The solution set is \(\left\{ \frac{7}{6} \right\}\), which corresponds to option A.
Examples
Logarithmic equations are used in fields like acoustics to calculate sound intensity levels in decibels, or in chemistry to determine pH levels of solutions. For example, if a chemist is monitoring the concentration of hydrogen ions in a reaction to ensure it stays within a specific safe range, they might encounter logarithmic expressions representing these concentrations. Understanding how to solve such equations allows researchers to determine the precise concentrations needed for a reaction to proceed safely and effectively.