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 = -2-32^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 = xx² + 3x = x(x + 3) = 0
This equation has two possible solutions:
Therefore, the solutions for the original logarithmic equation are x = 0 and x = -3.
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 = -2-3
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 = x
x² + 3x =
x(x + 3) = 0
This equation has two possible solutions:
x = 0x = -3Therefore, the solutions for the original logarithmic equation are x = 0 and x = -3.