Question

2^(x+2) + 2^(x+5) < 9

Answer 1

Объяснение:

$2^{x+2} + 2^{x+5} < 9\\ \\ 2^{x+2}+2^{x+2}*2^3$

Answer 2

To solve the inequality $2^{x+2} + 2^{x+5} < 9$, we can break it down into two separate inequalities and solve for $x$:

1. $2^{x+2} < 9 - 2^{x+5}$
2. $2^{x+5} < 9 - 2^{x+2}$

Let's start with the first inequality:

$2^{x+2} < 9 - 2^{x+5}$

We can subtract $2^{x+2}$ from both sides:

$0 < 9 - 2^{x+2} - 2^{x+5}$

Simplify the right side:

$0 < 9 - 2^{x+2} (1 + 2^3)$

$0 < 9 - 2^{x+2} - 8 \cdot 2^{x}$

Now, divide both sides by $2^{x}$ (since $2^{x}$ is always positive):

$0 < \frac{9}{2^{x}} - 2^2$

$0 < \frac{9}{2^{x}} - 4$

$4 < \frac{9}{2^{x}}$

Now, we can solve for $x$:

$\frac{4}{9} < \frac{1}{2^{x}}$

$2^{x} < \frac{9}{4}$

Taking the logarithm base 2 of both sides:

$x < \log_2 \left(\frac{9}{4}\right)$

$x < \log_2 2.25$

$x < \frac{\log 2.25}{\log 2}$

Using a calculator, we find:

$x < \frac{0.352}{0.301}$

$x < 1.17$

Now, let's move on to the second inequality:

$2^{x+5} < 9 - 2^{x+2}$

We can subtract $2^{x+5}$ from both sides:

$0 < 9 - 2^{x+5} - 2^{x+2}$

$0 < 9 - 2^{x+2} (2^3 - 1)$

$0 < 9 - 2^{x+2} - 7 \cdot 2^{x}$

Following similar steps as before, we eventually get:

$x < \frac{\log 1.125}{\log 2}$

$x < 0.15$

Now, combining the solutions from both inequalities:

$0.15 < x < 1.17$

So, the solution for the inequality $2^{x+2} + 2^{x+5} < 9$ is $0.15 < x < 1.17$.