Question

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

Answer 1

Заранее извиняюсь за почерк

Answer 2

To solve the equation log2(3x - 1) + log2(x - 1) = 1 + log2(x + 5), we can use logarithmic properties to simplify the equation and find the value of x.

Logarithmic property used:

1. log_a(b) + log_a(c) = log_a(b * c)

Applying the property, we can combine the two logarithms on the left side of the equation:

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

So the equation becomes:

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

Now, another logarithmic property allows us to convert the constant 1 on the right side into a logarithm:

2. log_a(b) = c is equivalent to b = a^c

Using this property, we can rewrite the equation as:

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

Now we simplify further:

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

Since 2^(log2(x + 5)) is just (x + 5), the equation becomes:

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

Now we have a quadratic equation, and we can solve for x:

Expand the left side:

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

Combine like terms:

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

Move all terms to one side to set the equation to zero:

3x^2 - 4x - 2x + 1 - 10 = 0

3x^2 - 6x - 9 = 0

Now we need to solve this quadratic equation. We can use the quadratic formula:

x = (-b ± √(b^2 - 4ac)) / 2a

where a = 3, b = -6, and c = -9

x = (6 ± √((-6)^2 - 4 * 3 * -9)) / 2 * 3

x = (6 ± √(36 + 108)) / 6

x = (6 ± √144) / 6

Now, calculate the two possible values of x:

x = (6 + 12) / 6 = 18 / 6 = 3

x = (6 - 12) / 6 = -6 / 6 = -1

Therefore, the equation log2(3x - 1) + log2(x - 1) = 1 + log2(x + 5) has two solutions: x = 3 and x = -1. However, it's essential to check these solutions in the original equation, as some solutions may not be valid due to the domain of the logarithmic functions. Always check the solutions and discard any that result in an invalid argument for the logarithms.