Question

Решить систему 2^x × 3^y = 108 2^x + 3^y = 31

Answer 1

Пошаговое объяснение:

Применим замену переменных и получим:

1) А * В = 108

2) А + В = 31 или В = 31 - А

3) А*(31 - А) = 108

4) А² - 31*А + 108 = 0 - решаем квадратное уравнение.

D = 529, √529 = 23

A₁ = 27, В₁ = 31 - 27 = 4

A₂ = 4, В₂ =31-4 = 27 - возвращаемся к подстановке

2ˣ = 4 = 2², $3^y = 27 = 3^3$,  х = 2, у = 3 - ответ

Есть и логарифмическое решение:

2ˣ = 27, х =㏒₂3³ = 3*㏒₂3 ≈ 3*1,584963 ≈ 4,7548875 =x - ответ.

$3^y=4$, y=㏒₃2² = 2*㏒₃2 ≈ 2*0.63093 ≈1.26186 =y - ответ

Answer 2

To solve the system of equations:

1. 2^x × 3^y = 108
2. 2^x + 3^y = 31

We can use the substitution method. Let's solve for one variable in terms of the other from one of the equations and then substitute it into the other equation.

Let's first solve equation (1) for 2^x in terms of y: 2^x = 108 / 3^y

Now, substitute this value for 2^x into equation (2): (108 / 3^y) + 3^y = 31

To simplify the equation, let's get rid of the fraction by multiplying everything by 3^y: 108 + (3^y)^2 = 31 * 3^y

Now, rearrange the equation: (3^y)^2 - 31 * 3^y + 108 = 0

This is a quadratic equation in terms of 3^y. Let's solve for 3^y using the quadratic formula:

The quadratic formula is given by: x = (-b ± √(b^2 - 4ac)) / 2a

In our equation, a = 1, b = -31, and c = 108. Plugging these values into the formula:

3^y = [31 ± √((-31)^2 - 4 * 1 * 108)] / 2 * 1

3^y = [31 ± √(961 - 432)] / 2

3^y = [31 ± √529] / 2

Now, we have two cases:

Case 1: 3^y = (31 + √529) / 2 = (31 + 23) / 2 = 54 / 2 = 27 Case 2: 3^y = (31 - √529) / 2 = (31 - 23) / 2 = 8 / 2 = 4

Now, let's solve for x using each case:

Case 1: 3^y = 27 2^x = 108 / 27 = 4 x = log2(4) = 2

Case 2: 3^y = 4 2^x = 108 / 4 = 27 x = log2(27) ≈ 4.7549

So, the solutions to the system of equations are: Case 1: x = 2 and y = 3 Case 2: x ≈ 4.7549 and y = 1

Please note that one of the solutions has fractional values, which might not be apparent due to rounding in intermediate steps.