Question

Log(2) (x+5)- log(2) (x-1) > 1

Answer 1

ОДЗ: x>-5, x>1. x∈(1;+∞)

log2((x+5)/(x-1))>1

log2((x+5)/(x-1))>log2(2)

(x+5)/(x-1)>2

(x+5-2x+2)/(x-1)>0

(7-x)/(x-1)>0

методом интервалов получаем ответ: x∈(1;7)

Answer 2

To solve the inequality $\log_{2}(x+5) - \log_{2}(x-1) > 1$, we can use logarithmic properties to simplify it. The properties we will use are:

1. $\log_{a}(b) - \log_{a}(c) = \log_{a}\left(\dfrac{b}{c}\right)$
2. If $\log_{a}(b) > c$, then $b > a^c$

Let's apply these properties step-by-step:

Step 1: Apply the first property to combine the logarithms:

$\log_{2}(x+5) - \log_{2}(x-1) = \log_{2}\left(\dfrac{x+5}{x-1}\right) > 1$

Step 2: Apply the second property:

$\dfrac{x+5}{x-1} > 2^1$

$\dfrac{x+5}{x-1} > 2$

Step 3: Now, let's solve for $x$ by isolating it on one side of the inequality. First, let's multiply both sides by $x-1$:

$x+5 > 2(x-1)$

Step 4: Distribute on the right side:

$x+5 > 2x - 2$

Step 5: Move $2x$ to the left side and $5$ to the right side:

$x - 2x > 5 - 2$

$-x > 3$

Step 6: To solve for $x$, multiply both sides by $-1$ (which reverses the inequality):

$x < -3$

So, the solution to the inequality is $x < -3$. However, when dealing with logarithmic inequalities, we need to be cautious about the domain of the logarithmic expression.

Recall that for a logarithm to be defined, the argument (what's inside the logarithm) must be greater than zero. So, $x+5$ and $x-1$ must be greater than zero:

$x+5 > 0 \implies x > -5$

$x-1 > 0 \implies x > 1$

Thus, the valid range for $x$ is $x > 1$ since we need both $x+5$ and $x-1$ to be positive.

Therefore, the final solution to the inequality is $1 < x < -3$.