Problem:
Given the equation:
[tex]\log_5(3a-5)=\log_5(-3a+1)[/tex]
Solution Steps:
- Equal Bases Property:
Since the logarithms have the same base, the arguments must be
equal for the equation to hold:
[tex]3a-5=-3a+1[/tex]
2. Solve for [tex]a[/tex]:
To solve this equation, first add [tex]3a[/tex] to both sides:
[tex]3a-5+3a=-3a+1+3a\\\text{or, }6a-5=1[/tex]
Next, add 5 to both sides:
[tex]6a-5+5=1+5\\\text{or, }6a=6[/tex]
Finally, divide both sides by 6:
[tex]a=1[/tex]
3. Check for Validity
Substitute [tex]a=1[/tex] back into the original logarithmic expression to
check if the result is valid.
[tex]\log_5(3.1-5)=\log_5(-2)\\\log_5(-3.1+1)=\log_5(-2)\\\\[/tex]
Both sides yield [tex]\log_5(-2).[/tex]
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 [tex]\log_5(-2)[/tex] is
undefined.
Conclusion:
Since the logarithms are not defined for the value of [tex]a=1[/tex], 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.
