Log₀,₂(-2-3x) = log₀,₂(x²-2)

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To solve this logarithmic equation, we first need to rewrite the equation in exponential form.

For the left side of the equation, we have:

Log₀,₂(-2-3x) = y is equivalent to 2^y = -2-3x

For the right side of the equation, we have:

Log₀,₂(x²-2) = y is equivalent to 2^y = x²-2

Now we can set these two exponential expressions equal to each other:

2^y = -2-3x
2^y = x²-2

Since both expressions are equal to 2^y, we can set them equal to each other:

-2-3x = x²-2

Now, let's solve for x:

-3x = x²
x² + 3x = 0
x(x + 3) = 0

This equation has two possible solutions:

Therefore, the solutions for the original logarithmic equation are x = 0 and x = -3.