Question

Log0.4x+log0.4(x-1)>log0.4(x+3)

Answer 1

Ответ:

(1;3)

Объяснение:

log0,4x+log0,4(x-1)>log0,4(x+3)

log0,4(x(x-1))>log0,5(x+3)

x^2 - x < x + 3

x^2 - 2x -3 < 0

x € (-1;3)

x-1>0 => x>1

x € (1;3)

Answer 2

To solve the inequality $\log_{0.4}(x) + \log_{0.4}(x-1) > \log_{0.4}(x+3)$, we need to apply the properties of logarithms and algebraic manipulation. First, we'll combine the logarithms on the left-hand side:

$\log_{0.4}(x) + \log_{0.4}(x-1) = \log_{0.4}(x \cdot (x-1)) = \log_{0.4}(x^2 - x)$.

Now, our inequality becomes:

$\log_{0.4}(x^2 - x) > \log_{0.4}(x+3)$.

To remove the logarithms, we can rewrite the inequality as:

$0.4^{(\log_{0.4}(x^2 - x))} > 0.4^{(\log_{0.4}(x+3))}$.

Since the base of the logarithm and the base of the exponential function are the same, the logarithm and exponential functions cancel each other out. So we get:

$x^2 - x > x + 3$.

Now, let's bring all the terms to one side of the inequality:

$x^2 - x - (x + 3) > 0$.

Simplifying further:

$x^2 - x - x - 3 > 0$.

$x^2 - 2x - 3 > 0$.

Now, we need to find the critical points where the expression equals zero:

$x^2 - 2x - 3 = 0$.

To factor this quadratic equation, we look for two numbers whose product is -3 and whose sum is -2. These numbers are -3 and +1:

$(x - 3)(x + 1) = 0$.

Setting each factor to zero:

$x - 3 = 0$ or $x + 1 = 0$.

Solving for $x$:

$x = 3$ or $x = -1$.

Now, we can test different intervals of $x$ to determine when the inequality is satisfied. You can use a sign chart or pick test points. I'll use a sign chart:

Interval	Test Value	Sign of $x^2 - 2x - 3$
$x < -1$	$x = -2$	Positive
$-1 < x < 3$	$x = 0$	Negative
$x > 3$	$x = 4$	Positive

From the sign chart, we can see that the inequality $x^2 - 2x - 3 > 0$ is true for $x < -1$ and $x > 3$. Thus, the solution to the original inequality $\log_{0.4}(x) + \log_{0.4}(x-1) > \log_{0.4}(x+3)$ is:

$x < -1$ or $x > 3$.