Log2(x-1)+log2(x+3)=1

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вопрос

To solve the equation, we can use the properties of logarithms to combine the two terms on the left side of the equation.

First, rewrite the given equation as a single logarithm:

log2[(x-1)(x+3)] = 1

Now, we can rewrite the equation in exponential form:

2^1 = (x-1)(x+3)

2 = (x-1)(x+3)

Expand the right side of the equation:

2 = x^2 + 2x - x - 3

2 = x^2 + x - 3

Rearrange the equation to set it equal to zero:

x^2 + x - 5 = 0

Now, we can solve this quadratic equation using the quadratic formula:

x = [-b ± sqrt(b^2 - 4ac)] / 2a

In this case, a = 1, b = 1, and c = -5. Plugging these values into the formula:

x = [-1 ± sqrt(1^2 - 41-5)] / 2*1
x = [-1 ± sqrt(1 + 20)] / 2
x = [-1 ± sqrt(21)] / 2

Therefore, the solutions to the equation are:

x = (-1 + √21)/2 or x = (-1 - √21)/2