3*4^х-7*2^х+1-5<0

To solve the inequality \( 3 \cdot 4^x - 7 \cdot 2^x + 1 - 5 < 0 \), let's first rewrite the expression in simpler terms.

Notice that \( 4^x = (2^2)^x = (2^x)^2 \). Let's set \( y = 2^x \). Then, we can express the inequality as:

\[ 3y^2 - 7y + (1 - 5) < 0 \]

This simplifies to:

\[ 3y^2 - 7y - 4 < 0 \]

Next, we need to find the roots of the quadratic equation \( 3y^2 - 7y - 4 = 0 \) using the quadratic formula:

\[ y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

Here, \( a = 3 \), \( b = -7 \), and \( c = -4 \). Plugging in these values:

\[ y = \frac{7 \pm \sqrt{(-7)^2 - 4 \cdot 3 \cdot (-4)}}{2 \cdot 3} \]

\[ y = \frac{7 \pm \sqrt{49 + 48}}{6} \]

\[ y = \frac{7 \pm \sqrt{97}}{6} \]

Now, we can denote the roots as:

\[ y_1 = \frac{7 + \sqrt{97}}{6}, \quad y_2 = \frac{7 - \sqrt{97}}{6} \]

Next, we observe that the quadratic \( 3y^2 - 7y - 4 \) opens upwards (since the coefficient of \( y^2 \) is positive). The inequality \( 3y^2 - 7y - 4 < 0 \) will be satisfied for values of \( y \) between the two roots:

\[ y_2 < y < y_1 \]

Substituting back \( y = 2^x \):

\[ \frac{7 - \sqrt{97}}{6} < 2^x < \frac{7 + \sqrt{97}}{6} \]

Now, we need the numerical values of the roots:

Calculating \( \sqrt{97} \) (approximately \( 9.8489 \)):

\[ y_1 \approx \frac{7 + 9.8489}{6} \approx \frac{16.8489}{6} \approx 2.808 \quad \text{(approximately)} \]

\[ y_2 \approx \frac{7 - 9.8489}{6} \approx \frac{-2.8489}{6} \approx -0.475 \quad \text{(approximately)} \]

Since \( y = 2^x \) must be positive, we disregard \( y_2 \):

We focus on:

\[ 2^x < \frac{7 + \sqrt{97}}{6} \approx 2.808 \]

Thus, we take the logarithm base 2:

\[ x < \log_2(2.808) \]

Calculating \( \log_2(2.808) \):

Using the change of base formula, \( \log_2(a) = \frac{\log_{10}(a)}{\log_{10}(2)} \):

\[ \log_2(2.808) \approx \frac{\log_{10}(2.808)}{0.301} \approx \frac{0.448}{0.301} \approx 1.486 \]

Final form of the solution, rounding off:

\[ x < 1.486 \]

Thus, the solution to the inequality \( 3 \cdot 4^x - 7 \cdot 2^x - 4 < 0 \) is:

\[ x < \log_2\left(\frac{7 + \sqrt{97}}{6}\right) \]

or approximately:

\[ x < 1.486 \]