Problem:
Given the equation:
\(\log_5(3a-5)=\log_5(-3a+1)\)
Solution Steps:
- Equal Bases Property:
Since the logarithms have the same base, the arguments must be
equal for the equation to hold:
\(3a-5=-3a+1\)
2. Solve for \(a\):
To solve this equation, first add \(3a\) to both sides:
\(3a-5+3a=-3a+1+3a\\\text{or, }6a-5=1\)
Next, add 5 to both sides:
\(6a-5+5=1+5\\\text{or, }6a=6\)
Finally, divide both sides by 6:
\(a=1\)
3. Check for Validity
Substitute \(a=1\) back into the original logarithmic expression to
check if the result is valid.
\(\log_5(3.1-5)=\log_5(-2)\\\log_5(-3.1+1)=\log_5(-2)\\\\\)
Both sides yield \(\log_5(-2).\)
4. Analyze the Argument:
The logarithm of a negative number is not defined in the set of
real numbers. Since -2 is negative, the expression \(\log_5(-2)\) is
undefined.
Conclusion:
Since the logarithms are not defined for the value of \(a=1\), there is no real solution to the given equation.
Complex solutions:
To consider complex solutions, let it be known that the logarithm function can be extended to complex numbers. But such analysis involves a more advanced approach involving complex process would be necessary, so it goes beyond typical real-number logarithm problems.
