Given:
Given the equation
[tex]\log_3(4-a)=\log_3(-2a+2)[/tex]
Required: Value of a
Explanation:
The given equation can be written as
[tex]\log_3(4-a)-\log_3(-2a+2)=0[/tex]
Use the property
[tex]\log_bp-\log_bq=\log_b\frac{p}{q}[/tex]
Thus,
[tex]\log_3(\frac{4-a}{-2a+2})=0[/tex]
Take antilogarithm on both sides.
[tex]\begin{gathered} \frac{4-a}{-2a+2}=3^0 \\ 4-a=-2a+2 \\ -a+2a=2-4 \\ a=-2 \end{gathered}[/tex]
Final Answer: The value of a is -2.
