Log2(3x-1) + log2(x-1) = 1 + log2( x+5)

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To solve this equation, we can combine the two logarithms on the left side using the product rule of logarithms:

log2(3x - 1) * (x - 1) = log2((x + 5)) + log2(2)

Simplify the left side:

log2((3x - 1)(x - 1)) = log2(2(x + 5))

Now, set the expressions inside the logarithms equal to each other:

(3x - 1)(x - 1) = 2(x + 5)

Expand both sides:

3x^2 - 3x - x + 1 = 2x + 10

3x^2 - 4x + 1 = 2x + 10

Rearrange to set the equation equal to zero:

3x^2 - 6x - 9 = 0

Now, we can solve this quadratic equation by factoring or using the quadratic formula. Factoring, we get:

3(x - 3)(x + 1) = 0

This gives us two possible solutions:

x = 3 or x = -1

To check these solutions, substitute them back into the original equation:

For x = 3:
log2(3(3) - 1) + log2(3 - 1) = 1 + log2(3 + 5)
log2(8) + log2(2) = 1 + log2(8)
3 + 1 = 1 + 3
4 = 4

For x = -1:
log2(3(-1) - 1) + log2(-1 - 1) = 1 + log2(-1 + 5)
log2(-4) is undefined as a logarithm with a negative argument is undefined.
Therefore, x = -1 is not a valid solution.

Hence, the solution to the equation is x = 3.