Solving the Equation log2(3-x) log2(1-x) 3

In this article, we will walk through the process of solving the equation log2(3-x) log2(1-x) 3. We will use the properties of logarithms and basic algebra to find the solution, ensuring that the arguments of the logarithms are positive.

Step-by-Step Solution

Let's start with the given equation:

log2(3-x) log2(1-x) 3

Step 1: Rewrite the Equation

First, let's rewrite the equation by combining the logarithms on the left side using the product property:

log2(3-x)1-x 3

According to the logarithm properties, this can be rewritten as:

3 - x1 - x 23

Step 2: Simplify the Equation

Calculate the right-hand side of the equation:

3 - x1 - x 8

Now, expand the left side of the equation:

3 - 3x - x x2 8

This simplifies to:

x2 - 4x - 3 8

Next, rearrange the equation to form a quadratic equation:

x2 - 4x - 5 0

Step 3: Factor the Quadratic Equation

Factor the quadratic equation:

x2 - 4x - 5 (x - 5)(x 1) 0

This gives us two solutions:

x - 5 0 rarr; x 5

x 1 0 rarr; x -1

Step 4: Check for Valid Solutions

Now we need to check which of these solutions are valid in the original logarithmic expressions. The arguments of the logarithms must be positive:

x 5 3 - 5 -2 (not valid) 1 - 5 -4 (not valid) x -1 3 - (-1) 4 (valid) 1 - (-1) 2 (valid)

Since x 5 is not a valid solution, the only valid solution is:

boxed{-1}

Alternative Solutions

Let's also present an alternative approach to solving this equation:

3 - x1 - x 23 8

This gives us:

3 - x1 - x - 8 0

Further simplifying:

x2 - 4x - 5 0

Factoring the quadratic equation:

x2 - 5x x - 5 0

Grouping the factors:

x(x - 5) 1(x - 5) 0

This can be factored as:

(x - 5)(x 1) 0

Setting each factor to zero:

x - 5 0 rarr; x 5

x 1 0 rarr; x -1

Verification

Verify that the solutions are valid:

For x 5:

3 - 5 -2 (not valid) 1 - 5 -4 (not valid)

For x -1:

3 - (-1) 4 (valid) 1 - (-1) 2 (valid)

Therefore, the only valid solution is:

boxed{-1}