Given the equation:
[tex]\log_5(3a-5)=\log_5(-3a+1)[/tex]
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.
Since the logarithms are not defined for the value of [tex]a=1[/tex], there is no real solution to the given equation.
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.