Question

Решите задание номер 3

Answer 1

Задание-то где?
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Answer 2

Task 3: Solving Equations

To solve the given equations, let's go through each one step by step.

1. \(log_3(x^2+4)=log_3(4x)\)

To solve this equation, we can use the property of logarithms that states that if \(log_a(b) = log_a(c)\), then \(b = c\). Applying this property to the given equation, we have:

\(x^2 + 4 = 4x\)

Rearranging the equation, we get:

\(x^2 - 4x + 4 = 0\)

This is a quadratic equation. To solve it, we can factorize it:

\((x - 2)^2 = 0\)

Taking the square root of both sides, we have:

\(x - 2 = 0\)

Solving for \(x\), we get:

\(x = 2\)

Therefore, the solution to the equation is \(x = 2\).

2. \(log_28=log_{16}(x)+2\)

To solve this equation, we can use the property of logarithms that states that if \(log_a(b) = log_a(c)\), then \(b = c\). Applying this property to the given equation, we have:

\(8 = 16^{\left(x+2\right)}\)

We can rewrite \(16\) as \(2^4\) since \(16\) is equal to \(2\) raised to the power of \(4\):

\(8 = \left(2^4\right)^{\left(x+2\right)}\)

Using the property of exponents that states that \(\left(a^b\right)^c = a^{b \cdot c}\), we can simplify the equation:

\(8 = 2^{4 \cdot \left(x+2\right)}\)

Simplifying further, we have:

\(8 = 2^{4x+8}\)

Since both sides of the equation have the same base, we can equate the exponents:

\(4x + 8 = 3\)

Solving for \(x\), we get:

\(4x = -5\)

\(x = -\frac{5}{4}\)

Therefore, the solution to the equation is \(x = -\frac{5}{4}\).

3. \(log_2(2x^2)-5=log_2(x) +log_2(x-5)\)

To solve this equation, we can simplify the logarithmic expressions using the properties of logarithms. First, let's simplify the left side of the equation:

\(log_2(2x^2)-5 = log_2(x) + log_2(x-5)\)

Using the property of logarithms that states that \(log_a(b) - log_a(c) = log_a\left(\frac{b}{c}\right)\), we can rewrite the equation as:

\(log_2\left(\frac{2x^2}{2^5}\right) = log_2\left(\frac{x(x-5)}{2}\right)\)

Simplifying further, we have:

\(log_2\left(\frac{x^2}{32}\right) = log_2\left(\frac{x(x-5)}{2}\right)\)

Using the property of logarithms that states that if \(log_a(b) = log_a(c)\), then \(b = c\), we can equate the expressions inside the logarithms:

\(\frac{x^2}{32} = \frac{x(x-5)}{2}\)

Multiplying both sides of the equation by \(32\) to eliminate the fraction, we have:

\(x^2 = 16x(x-5)\)

Expanding and rearranging the equation, we get:

\(x^2 = 16x^2 - 80x\)

\(15x^2 - 80x = 0\)

Factoring out \(x\), we have:

\(x(15x - 80) = 0\)

Setting each factor equal to zero, we get two possible solutions:

\(x = 0\) or \(x = \frac{80}{15}\)

Simplifying the second solution, we have:

\(x = \frac{16}{3}\)

Therefore, the solutions to the equation are \(x = 0\) and \(x = \frac{16}{3}\).

4. \(log_9(x+4)\geq log_9(2x)^2\)

To solve this equation, we can use the property of logarithms that states that if \(log_a(b) \geq log_a(c)\), then \(b \geq c\). Applying this property to the given equation, we have:

\(x + 4 \geq (2x)^2\)

Simplifying the right side of the equation, we have:

\(x + 4 \geq 4x^2\)

Rearranging the equation, we get:

\(4x^2 - x - 4 \leq 0\)

To solve this quadratic inequality, we can factorize it:

\((2x - 1)(2x + 4) \leq 0\)

To determine the sign of the expression, we can use a sign chart or test values. Testing values in each interval, we find that the solution is:

\(-\frac{4}{3} \leq x \leq 1\)

Therefore, the solution to the inequality is \(-\frac{4}{3} \leq x \leq 1\).

5. \(log_{500}500\geq log_2(1+3x)\)

To solve this equation, we can simplify the logarithmic expressions using the properties of logarithms. First, let's simplify the left side of the equation:

\(log_{500}500 \geq log_2(1+3x)\)

Using the property of logarithms that states that if \(log_a(b) \geq log_a(c)\), then \(b \geq c\), we can rewrite the equation as:

\(500 \geq 2^{1+3x}\)

Simplifying further, we have:

\(500 \geq 2 \cdot 2^{3x}\)

Dividing both sides of the equation by \(2\), we get:

\(250 \geq 2^{3x}\)

To solve for \(x\), we can take the logarithm of both sides of the equation:

\(log_2(250) \geq log_2(2^{3x})\)

Using the property of logarithms that states that \(log_a(b^c) = c \cdot log_a(b)\), we can simplify the equation:

\(log_2(250) \geq 3x \cdot log_2(2)\)

Since \(log_2(2) = 1\), we have:

\(log_2(250) \geq 3x\)

Dividing both sides of the equation by \(3\), we get:

\(\frac{log_2(250)}{3} \geq x\)

Therefore, the solution to the inequality is \(x \leq \frac{log_2(250)}{3}\).

In summary, the solutions to the given equations are as follows: 1. \(x = 2\) 2. \(x = -\frac{5}{4}\) 3. \(x = 0\) and \(x = \frac{16}{3}\) 4. \(-\frac{4}{3} \leq x \leq 1\) 5. \(x \leq \frac{log_2(250)}{3}\)