(2^x+4)^x-3=0,5^x•4^x-4

Question 1

Задача Задача №2170598

А

Аноним

10 ноября 2024 18:06 · публичный вопрос

А

Аноним 10 ноября 2024 18:06

(2^x+4)^x-3=0,5^x•4^x-4

Answer

To solve the equation \( (2^x + 4)^x - 3 = 5^x \cdot 4^x - 4 \), we will start by simplifying both sides.

First, notice that \( 4 = 2^2 \), so we can rewrite \( 5^x \cdot 4^x \) as follows:

\[ 5^x \cdot 4^x = 5^x \cdot (2^2)^x = 5^x \cdot 2^{2x} = 5^x \cdot 2^{2x}. \]

Now, we simplify the original equation:

\[ (2^x + 4)^x - 3 = 5^x \cdot 2^{2x} - 4. \]

This equation isn't easily solvable algebraically, so we might benefit from trying to find particular values of \( x \).

Let’s consider \( x = 2 \):

For the left-hand side: \[ (2^2 + 4)^2 - 3 = (4 + 4)^2 - 3 = 8^2 - 3 = 64 - 3 = 61. \]
For the right-hand side: \[ 5^2 \cdot 4^2 - 4 = 25 \cdot 16 - 4 = 400 - 4 = 396. \]

Since \( 61 \neq 396\), \( x=2 \) is not a solution.

Let’s try \( x = 1 \):

For the left-hand side: \[ (2^1 + 4)^1 - 3 = (2 + 4)^1 - 3 = 6 - 3 = 3. \]
For the right-hand side: \[ 5^1 \cdot 4^1 - 4 = 5 \cdot 4 - 4 = 20 - 4 = 16. \]

Since \( 3 \neq 16\), \( x=1 \) is also not a solution.

Next, let's try \( x = 0 \):

For the left-hand side: \[ (2^0 + 4)^0 - 3 = (1 + 4)^0 - 3 = 1 - 3 = -2. \]
For the right-hand side: \[ 5^0 \cdot 4^0 - 4 = 1 \cdot 1 - 4 = 1 - 4 = -3. \]

Since \( -2 \neq -3\), \( x=0 \) is not a solution.

Now, let's try \( x = -1 \):

For the left-hand side: \[ (2^{-1} + 4)^{-1} - 3 = \left(\frac{1}{2} + 4\right)^{-1} - 3 = \left(\frac{1}{2} + \frac{8}{2}\right)^{-1} - 3 = \left(\frac{9}{2}\right)^{-1} - 3 = \frac{2}{9} - 3 = \frac{2 - 27}{9} = \frac{-25}{9}. \]
For the right-hand side: \[ 5^{-1} \cdot 4^{-1} - 4 = \frac{1}{5} \cdot \frac{1}{4} - 4 = \frac{1}{20} - 4 = \frac{1}{20} - \frac{80}{20} = \frac{-79}{20}. \]

Since \( \frac{-25}{9} \neq \frac{-79}{20}\), \( x=-1 \) is also not a solution.

Now, checking integers in negative space we can try \( x = -2 \):

For the left-hand side: \[ (2^{-2} + 4)^{-2} - 3 = \left(\frac{1}{4} + 4\right)^{-2} - 3 = \left(\frac{1}{4} + \frac{16}{4}\right)^{-2} - 3 = \left(\frac{17}{4}\right)^{-2} - 3 = \frac{16}{289} - 3 = \frac{16 - 867}{289} = \frac{-851}{289}. \]
For the right-hand side: \[ 5^{-2} \cdot 4^{-2} - 4 = \left(\frac{1}{25} \times \frac{1}{16}\right) - 4 = \frac{1}{400} - 4 = \frac{1}{400} - \frac{1600}{400} = \frac{-1599}{400}. \]

Since \( \frac{-851}{289} \neq \frac{-1599}{400}\), \( x=-2 \) is not a solution as well.

Trying \( x = -3 \):

For the left-hand side: \[ (2^{-3} + 4)^{-3} - 3 = \left(\frac{1}{8} + 4\right)^{-3} - 3 = \left(\frac{1}{8} + \frac{32}{8}\right)^{-3} - 3 = \left(\frac{33}{8}\right)^{-3} - 3 = \frac{512}{35937} - 3 = \frac{512 - 107811}{35937} = \frac{-107299}{35937}. \]
For the right-hand side: \[ 5^{-3} \cdot 4^{-3} - 4 = \left(\frac{1}{125} \cdot \frac{1}{64}\right) - 4 = \frac{1}{8000} - 4 = \frac{1 - 32000}{8000} = \frac{-31999}{8000}. \]

Since \( \frac{-107299}{35937} \neq \frac{-31999}{8000}\), that value isn't a solution either.

Another approach might require numerical or graphical solutions, as algebraically it's becoming increasingly complex to find a solvable value for \( x \).

Let me know if you would like to go and calculate or graph the functions involved to find a solution.

Question 2

(2^x+4)^x-3=0,5^x•4^x-4

Answer

To solve the equation \( (2^x + 4)^x - 3 = 5^x \cdot 4^x - 4 \), we will start by simplifying both sides. First, notice that \( 4 = 2^2 \), so we can rewrite \( 5^x \cdot 4^x \) as follows: \[ 5^x \cdot 4^x = 5^x \cdot (2^2)^x = 5^x \cdot 2^{2x} = 5^x \cdot 2^{2x}. \] Now, we simplify the original equation: \[ (2^x + 4)^x - 3 = 5^x \cdot 2^{2x} - 4. \] This equation isn't easily solvable algebraically, so we might benefit from trying to find particular values of \( x \). Let’s consider \( x = 2 \): - For the left-hand side: \[ (2^2 + 4)^2 - 3 = (4 + 4)^2 - 3 = 8^2 - 3 = 64 - 3 = 61. \] - For the right-hand side: \[ 5^2 \cdot 4^2 - 4 = 25 \cdot 16 - 4 = 400 - 4 = 396. \] Since \( 61 \neq 396\), \( x=2 \) is not a solution. Let’s try \( x = 1 \): - For the left-hand side: \[ (2^1 + 4)^1 - 3 = (2 + 4)^1 - 3 = 6 - 3 = 3. \] - For the right-hand side: \[ 5^1 \cdot 4^1 - 4 = 5 \cdot 4 - 4 = 20 - 4 = 16. \] Since \( 3 \neq 16\), \( x=1 \) is also not a solution. Next, let's try \( x = 0 \): - For the left-hand side: \[ (2^0 + 4)^0 - 3 = (1 + 4)^0 - 3 = 1 - 3 = -2. \] - For the right-hand side: \[ 5^0 \cdot 4^0 - 4 = 1 \cdot 1 - 4 = 1 - 4 = -3. \] Since \( -2 \neq -3\), \( x=0 \) is not a solution. Now, let's try \( x = -1 \): - For the left-hand side: \[ (2^{-1} + 4)^{-1} - 3 = \left(\frac{1}{2} + 4\right)^{-1} - 3 = \left(\frac{1}{2} + \frac{8}{2}\right)^{-1} - 3 = \left(\frac{9}{2}\right)^{-1} - 3 = \frac{2}{9} - 3 = \frac{2 - 27}{9} = \frac{-25}{9}. \] - For the right-hand side: \[ 5^{-1} \cdot 4^{-1} - 4 = \frac{1}{5} \cdot \frac{1}{4} - 4 = \frac{1}{20} - 4 = \frac{1}{20} - \frac{80}{20} = \frac{-79}{20}. \] Since \( \frac{-25}{9} \neq \frac{-79}{20}\), \( x=-1 \) is also not a solution. Now, checking integers in negative space we can try \( x = -2 \): - For the left-hand side: \[ (2^{-2} + 4)^{-2} - 3 = \left(\frac{1}{4} + 4\right)^{-2} - 3 = \left(\frac{1}{4} + \frac{16}{4}\right)^{-2} - 3 = \left(\frac{17}{4}\right)^{-2} - 3 = \frac{16}{289} - 3 = \frac{16 - 867}{289} = \frac{-851}{289}. \] - For the right-hand side: \[ 5^{-2} \cdot 4^{-2} - 4 = \left(\frac{1}{25} \times \frac{1}{16}\right) - 4 = \frac{1}{400} - 4 = \frac{1}{400} - \frac{1600}{400} = \frac{-1599}{400}. \] Since \( \frac{-851}{289} \neq \frac{-1599}{400}\), \( x=-2 \) is not a solution as well. Trying \( x = -3 \): - For the left-hand side: \[ (2^{-3} + 4)^{-3} - 3 = \left(\frac{1}{8} + 4\right)^{-3} - 3 = \left(\frac{1}{8} + \frac{32}{8}\right)^{-3} - 3 = \left(\frac{33}{8}\right)^{-3} - 3 = \frac{512}{35937} - 3 = \frac{512 - 107811}{35937} = \frac{-107299}{35937}. \] - For the right-hand side: \[ 5^{-3} \cdot 4^{-3} - 4 = \left(\frac{1}{125} \cdot \frac{1}{64}\right) - 4 = \frac{1}{8000} - 4 = \frac{1 - 32000}{8000} = \frac{-31999}{8000}. \] Since \( \frac{-107299}{35937} \neq \frac{-31999}{8000}\), that value isn't a solution either. Another approach might require numerical or graphical solutions, as algebraically it's becoming increasingly complex to find a solvable value for \( x \). Let me know if you would like to go and calculate or graph the functions involved to find a solution.