7^(x-y)=sqrt(7) log(x+3)=2+log(4*y-1)
Решение
Вы ввели
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x - y ___ 7 = \/ 7
$$7^{x - y} = \sqrt{7}$$
log(x + 3) = 2 + log(4*y - 1)
$$\log{\left (x + 3 \right )} = \log{\left (4 y - 1 \right )} + 2$$
Быстрый ответ
$$x_{1} = \frac{3 + 3 e^{2}}{-1 + 4 e^{2}}$$
=
$$\frac{3 + 3 e^{2}}{-1 + 4 e^{2}}$$
=
$$y_{1} = \frac{7 + 2 e^{2}}{-2 + 8 e^{2}}$$
=
$$\frac{7 + 2 e^{2}}{-2 + 8 e^{2}}$$
=
=
$$\frac{3 + 3 e^{2}}{-1 + 4 e^{2}}$$
=
0.881319881369251
$$y_{1} = \frac{7 + 2 e^{2}}{-2 + 8 e^{2}}$$
=
$$\frac{7 + 2 e^{2}}{-2 + 8 e^{2}}$$
=
0.381319881369251
Численный ответ
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x1 = 0.8813198813692509 y1 = 0.3813198813692509
x2 = 0.8813198813693521 + 1.306959156887441e-13*i y2 = 0.3813198813693521 + 1.306959156887441e-13*i
x3 = 0.8813198813692508 - 4.975276968287706e-17*i y3 = 0.3813198813692508 - 4.975276968287706e-17*i
x4 = 0.8813198813692509 y4 = 0.3813198813692509