Question

SOS (2^(x+5)-2^(-x))/(2^(3-x)-4^(-x))>2^x

Answer 1

Решение задания приложено

Answer 2

To solve the inequality $\frac{{2^{x+5}-2^{-x}}}{{2^{3-x}-4^{-x}}} > 2^x$, we'll need to manipulate the expression to isolate $x$ on one side. Let's go through the steps:

Step 1: Simplify the denominator Recall that $4^{-x} = \frac{1}{{4^x}}$ and $2^3 = 8$.

$\frac{{2^{x+5}-2^{-x}}}{{2^{3-x}-4^{-x}}} > 2^x$

$\frac{{2^{x+5}-2^{-x}}}{{2^{3-x}-\frac{1}{{4^x}}}} > 2^x$

Step 2: Get rid of fractions by multiplying both sides by $2^{3-x}-\frac{1}{{4^x}}$.

$(2^{x+5}-2^{-x}) > 2^x \cdot (2^{3-x}-\frac{1}{{4^x}})$

Step 3: Distribute $2^x$ on the right side.

$(2^{x+5}-2^{-x}) > 2^{x+3}-2^{-x}$

Step 4: Move the $2^{-x}$ term to the right side.

$2^{x+5} > 2^{x+3}$

Step 5: Divide both sides by $2^{x+3}$. Remember, we're assuming $2^{x+3} \neq 0$ since dividing by zero is undefined.

$\frac{{2^{x+5}}}{{2^{x+3}}} > 1$

Step 6: Subtract the exponents on the left side.

$2^{(x+5)-(x+3)} > 1$

$2^2 > 1$

Step 7: Simplify the left side.

$4 > 1$

The inequality $4 > 1$ is true for all values of $x$.

Therefore, the solution to the inequality $\frac{{2^{x+5}-2^{-x}}}{{2^{3-x}-4^{-x}}} > 2^x$ is all real values of $x$. In other words, it holds true for any value of $x$.