Question

Apply properties of logarithms to solve exponential and logarithmic equations

Answer 1

Given:

Given the equation

`\log_3(4-a)=\log_3(-2a+2)`

Required: Value of a

Step-by-step explanation:

The given equation can be written as

`\log_3(4-a)-\log_3(-2a+2)=0`

Use the property

`\log_bp-\log_bq=\log_b(p)/(q)`

Thus,

`\log_3((4-a)/(-2a+2))=0`

Take antilogarithm on both sides.

`\begin{gathered} (4-a)/(-2a+2)=3^0 \\ 4-a=-2a+2 \\ -a+2a=2-4 \\ a=-2 \end{gathered}`

Final Answer: The value of a is -2.