Question

3 * 4^x - 2 * 9^x + 6^x < 0

Answer 1

Решение:

Все слагаемые разделим на 6^x > 0;

3* 4^x / 6^x + 2*9^x / 6^x - 5* 6^x / 6^x < 0;

3 * (4/6)^x + 2* (9/6)^x - 5 *1 < 0;

3*(2/3)^x + 2 * (3/2)^x - 5 < 0;

(2/3)^x = t > 0; (3/2)^t = 1 / t ;

3 * t + 2 / t - 5 < 0; * t ≠ 0;

(3t^2 + 2 - 5t) / t < 0;

(3t^2 - 5 t + 2) / t < 0;

t > 0; ⇒ 3 t^2 - 5t + 2 < 0

t1 = 1; t 2 = 2/3;

3(t - 1)*(t - 2/3) <0;

используем метод интервалов

+ - +

(0)------(2/3)-------(1)---------- t

при t > 0; ⇒ t ∈ (2/3; 1);

составим двойное неравенство :

2/3 < (2/3)^x < 1;

(2/3)^1 < (2/3)^x < (2/3)^0;

2/3 < 1; ⇒ 0 < x < 1.

Ответ :

х∈ (0; 1)

Answer 2

To solve the inequality $3 \cdot 4^x - 2 \cdot 9^x + 6^x < 0$, we need to find the values of $x$ that satisfy this inequality. Let's break down the steps:

1. Factor out common terms: $3 \cdot 4^x - 2 \cdot 9^x + 6^x = 3 \cdot 2^{2x} - 2 \cdot 3^{2x} + 6^x$.

2. Substitute $6^x$ with $2^{x} \cdot 3^{x}$: $3 \cdot 2^{2x} - 2 \cdot 3^{2x} + 2^x \cdot 3^x$.

3. Factor out $2^x$: $2^x \cdot (3 \cdot 2^{x} - 2 \cdot 3^{2x} + 3^x)$.

Now we have a product of terms involving $2^x$. To analyze the sign of this expression, we can examine the signs of each factor:

• $2^x$ is always positive for real values of $x$.
• The term $(3 \cdot 2^{x} - 2 \cdot 3^{2x} + 3^x)$ is more complex and needs further analysis.

We need to determine the conditions under which the second term is negative. To do this, let's look at the term $3 \cdot 2^{x} - 2 \cdot 3^{2x} + 3^x$ in isolation and find its critical points. We'll analyze the behavior of this term for different values of $x$:

• As $x$ becomes very large, $3^{2x}$ grows much faster than $2^x$, so the term becomes negative.
• As $x$ becomes very small (approaching negative infinity), $3^{2x}$ tends to infinity faster than $2^x$, making the term positive.
• For some value of $x$, the term will transition from positive to negative. This is the point we're interested in.

Let's find the value of $x$ for which the term $3 \cdot 2^{x} - 2 \cdot 3^{2x} + 3^x$ transitions from positive to negative:

$3 \cdot 2^{x} - 2 \cdot 3^{2x} + 3^x = 0$
$3 \cdot 2^{x} + 3^x = 2 \cdot 3^{2x}$
$3^x (2^{x} + 1) = 2 \cdot 3^{2x}$
$2^{x} + 1 = 2 \cdot 3^{x}$
$2^{x} = 2 \cdot 3^{x} - 1$.

Solving this equation numerically for $x$ will give us the approximate point at which the transition occurs. Once we have that point, we can determine the intervals of $x$ values for which the original inequality $3 \cdot 4^x - 2 \cdot 9^x + 6^x < 0$ is satisfied.

Please note that finding the exact analytical solution for $x$ might not be feasible, and numerical methods or graphing tools might be needed for accurate solutions.