Question

Решите неравенство: 4^x - 14 × 2^x - 32 > 0

Answer 1

Решение приложено к снимку:

Answer 2

To solve the inequality $4^x - 14 \times 2^x - 32 > 0$, we can use substitution to simplify the expression.

Let's make a substitution: $2^x = y$. Now the inequality becomes:

$4^x - 14 \times y - 32 > 0$

Rewriting $4^x$ in terms of $y$:

$(2^2)^x - 14 \times y - 32 > 0$

Using the property $(a^m)^n = a^{m \times n}$, we get:

$2^{2x} - 14 \times y - 32 > 0$

Now, substitute $y = 2^x$ back into the inequality:

$2^{2x} - 14 \times 2^x - 32 > 0$

The inequality is now in terms of a single variable, $x$, which is easier to solve. Let's factor it:

$2^x(2^x - 14) - 32 > 0$

Now, we have two cases to consider:

1. $2^x > 0$: This is always true because $2^x$ is positive for all real values of $x$.
2. $2^x - 14 > 0$

Solving the second case:

$2^x - 14 > 0$

Add 14 to both sides:

$2^x > 14$

Now, take the logarithm (base 2) of both sides to solve for $x$:

$x > \log_2(14)$

Using a calculator, we find:

$x > 3.807$

So the solution to the inequality is $x > 3.807$.

In interval notation, the solution is $x \in (3.807, +\infty)$.