Can someone please help me with problem number 50 in this algebra 2 assignment? The answer is provided right below it. Please show your work.

Directions: Solve the equation.



Can someone please help me with problem number 50 in this algebra 2 assignment The answer is provided right below it Please show your work Directions Solve the class=


Answer :

Problem:

Given the equation:

[tex]\log_5(3a-5)=\log_5(-3a+1)[/tex]

Solution Steps:

  1. 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.