log3(x-2)=2-log2(x+6)

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вопрос

To solve this logarithmic equation, we need to first combine the logarithms on the right side of the equation:

log3(x-2) = 2 - log2(x+6)

Using the properties of logarithms, we can rewrite the right side of the equation as a single logarithm:

log3(x-2) + log2(x+6) = 2

Next, we can use the property of logarithms that states log_a (b) + log_a (c) = log_a (b * c) to combine the two logarithms on the left side:

log3((x-2)(x+6)) = 2

Now, we have a single logarithm on the left side. To solve for x, we need to rewrite the equation in exponential form:

3^2 = (x-2)(x+6)

9 = x^2 + 4x - 12

Rearranging the equation to make it easier to solve, we get:

x^2 + 4x - 21 = 0

Now, we need to factor the quadratic equation or use the quadratic formula to solve for x. Factoring the equation, we get:

(x + 7)(x - 3) = 0

This gives us two possible solutions for x:

x = -7 or x = 3

However, we need to check these solutions in the original equation, as taking the logarithm of a negative number or zero is undefined. We find that x = -7 does not satisfy the original equation, so the only solution is:

x = 3