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

Explanation:

The given equation can be written as

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

Use the property

$\log_bp-\log_bq=\log_b\frac{p}{q}$

Thus,

$\log_3(\frac{4-a}{-2a+2})=0$

Take antilogarithm on both sides.

$\begin{gathered} \frac{4-a}{-2a+2}=3^0 \\ 4-a=-2a+2 \\ -a+2a=2-4 \\ a=-2 \end{gathered}$

Final Answer: The value of a is -2.