Log^2*3x = log^2*4 +log^2*6

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вопрос

To solve this equation, we will first apply the properties of logarithms to combine the terms on the right side of the equation:

log^23x - log^24 - log^2*6 = 0

Now, we can combine the logarithms using the property that states log(a) - log(b) = log(a/b):

log^2(3x / 4 6) = 0

Now, we simplify the expression in parentheses:

3x / (4 * 6) = 3x / 24 = x / 8

So, the equation becomes:

log^2*(x / 8) = 0

Since the base of the logarithm is not specified, we assume it is 10. Therefore, we can rewrite the equation as:

log(x / 8) = 0

Now, using the definition of logarithms, we can rewrite the equation as:

x / 8 = 10^0

x / 8 = 1

Multiplying both sides by 8, we get:

x = 8

Therefore, the solution to the equation is x = 8.