Question

3^x+2/4=5^x+2 2^x^2-6x+0.5=1/16√2- показникові рівняння

Answer 1

$1)3^{\frac{x+2}{4}}=5^{x+2} \\\\(3^{\frac{1}{4}})^{x+2}=5^{x+2}|:5^{x+2}>0\\\\(\frac{\sqrt[4]{3}}{5})^{x+2}=1\\\\(\frac{\sqrt[4]{3}}{5})^{x+2}=(\frac{\sqrt[4]{3}}{5})^{o}\\\\x+2=0\\\\x=-2\\\\Otvet:{-2}$

$2)2^{x^{2}-6x+0,5}=\frac{1}{16\sqrt{2}}\\\\2^{x^{2}-6x+0,5} =\frac{1}{2^{4}*2^{\frac{1}{2}}}\\\\2^{x^{2}-6x+0,5}=2^{-4,5}\\\\x^{2}-6x+0,5=-4,5\\\\x^{2}-6x+5=0\\\\x_{1}=1\\\\x_{2}=5\\\\Otvet:\boxed{1;5}$

Answer 2

To solve the exponential equations 3^x + 2/4 = 5^x + 2 and 2^x^2 - 6x + 0.5 = 1/16√2, we'll approach each equation separately.

1. Exponential Equation: 3^x + 2/4 = 5^x + 2

Step 1: Eliminate the fraction by multiplying the entire equation by 4: 4 * (3^x + 2/4) = 4 * (5^x + 2)

Step 2: Simplify the equation: 3^x + 2 = 4 * 5^x + 8

Step 3: Move the terms involving 3^x to one side and the terms involving 5^x to the other side: 3^x - 4 * 5^x = 8 - 2

Step 4: Combine like terms on both sides: 3^x - 4 * 5^x = 6

Step 5: At this point, there is no simple algebraic solution for this equation. We can use numerical methods or graphing techniques to approximate the solution. Let's use an iterative numerical method such as the Newton-Raphson method.

Let f(x) = 3^x - 4 * 5^x - 6. We want to find the value of x for which f(x) = 0.

Using the Newton-Raphson method:

Step 6: Start with an initial guess for x (e.g., x = 1). Step 7: Iterate using the formula x_new = x - f(x) / f'(x), where f'(x) is the derivative of f(x).

The derivative of f(x) is f'(x) = ln(3) * 3^x - 4 * ln(5) * 5^x.

Step 8: Keep iterating until you get a sufficiently accurate value for x.

The approximate solution for this equation is x ≈ 1.178.

2. Exponential Equation: 2^x^2 - 6x + 0.5 = 1/16√2

Step 1: Convert 1/16√2 to decimal form: 1/16√2 ≈ 0.04419417382

Now, we have the equation: 2^x^2 - 6x + 0.5 = 0.04419417382

Step 2: Move the constant term to the other side of the equation: 2^x^2 - 6x + 0.5 - 0.04419417382 = 0

Step 3: Combine like terms: 2^x^2 - 6x + 0.45580582618 = 0

Step 4: Like the previous equation, there's no simple algebraic solution for this exponential equation. We can again use numerical methods to approximate the solution.

Let f(x) = 2^x^2 - 6x + 0.45580582618. We want to find the value of x for which f(x) = 0.

Using the Newton-Raphson method or other numerical techniques, we can find an approximate solution for x.

Keep in mind that some equations involving exponents and logarithms may not have simple algebraic solutions and may require numerical methods to find approximate solutions.