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

Предыдущий
вопрос Следующий
вопрос

To solve this logarithmic equation, we can use the properties of logarithms, specifically the product rule which states that log(a) + log(b) = log(ab), and the fact that log(a) = log(b) if and only if a = b.

Given equation: log3(x - 2) + log3(x + 2) = log3(2x - 1)

Using the product rule, we can rewrite the left side of the equation as a single logarithm:

log3((x - 2)(x + 2)) = log3(2x - 1)

Now we have:

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

Expanding the left side:

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

Rearranging the equation:

x^2 - 2x - 3 = 0

Now, we have a quadratic equation that can be factored:

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

This gives us two possible solutions for x:

x = 3 or x = -1

However, we need to check these solutions in the original equation since taking the logarithm of a negative number is undefined.

For x = 3:

log3(3 - 2) + log3(3 + 2) = log3(2*3 - 1)
log3(1) + log3(5) = log3(6 - 1)
0 + log3(5) = log3(5)
log3(5) = log3(5)

For x = -1:

log3(-1 - 2) + log3(-1 + 2) = log3(2*(-1) - 1)
log3(-3) + log3(1) = log3(-2 - 1)

Taking the logarithm of a negative number is not defined, so x = -1 is not a valid solution.

Therefore, the only solution to the equation is x = 3.