Question

Log2(x+2) / log2(x-1) < 2

Answer 1

Решение на фотографии

Answer 2

To solve the inequality $\frac{\log_{2}(x+2)}{\log_{2}(x-1)} < 2$, we need to be careful with the domain of the logarithmic functions. The domain of the logarithm requires that the argument of the logarithm must be positive, so we have two cases to consider for this inequality:

Case 1: $x > 1$ (since $x - 1 > 0$) Case 2: $x+2 > 0 \Rightarrow x > -2$

Now, let's proceed with each case separately:

Case 1: $x > 1$ In this case, both $\log_{2}(x+2)$ and $\log_{2}(x-1)$ are defined, and we can manipulate the inequality as follows:

$\frac{\log_{2}(x+2)}{\log_{2}(x-1)} < 2$

$\log_{2}(x+2) < 2\log_{2}(x-1)$

Using the property of logarithms that $\log_{a}(b) = c$ is equivalent to $a^c = b$, we can rewrite the inequality as:

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

Simplify further:

$4x - 4 > x + 2$

$4x - x > 2 + 4$

$3x > 6$

$x > 2$

Thus, for Case 1, the solution is $x > 2$.

Case 2: $x > -2$ In this case, both $\log_{2}(x+2)$ and $\log_{2}(x-1)$ are still defined, and we can manipulate the inequality similarly:

$\frac{\log_{2}(x+2)}{\log_{2}(x-1)} < 2$

$\log_{2}(x+2) < 2\log_{2}(x-1)$

Using the property of logarithms again:

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

Simplify further:

$4x - 4 > x + 2$

$4x - x > 2 + 4$

$3x > 6$

$x > 2$

Since the solution $x > 2$ is valid in both cases, the combined solution to the inequality is $x > 2$.