Question

Lоg0, 1 (x ^ 2 + 2x - 3) < log0, 1 (x - 1) ;

Answer 1

Ответ:

Объяснение:

$log_{0,1}(x^{2} +2x-3) < log_{0,1}(x-1)\\x^{2} +2x-3 < x-1\\x^{2} +2x-3-x+1 < 0\\x^{2} +x-2 < 0\\x^{2} -x+2x-2 < 0\\x(x-1)+2(x-1) < 0\\(x+2)(x-1) < 0\\x+2 < 0\\x-1 < 0\\x_{1} < -2\\x_{2} < 1$

x ∈ (-2; 1)

Answer 2

The expression you provided contains a logarithm with a base of 0.1, which is not defined. Logarithms are typically defined with a positive base greater than 1, so this expression doesn't make mathematical sense.

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