16^x=64 (уравнение) Учитель очень удивится увидев твоё верное решение 😼
Найду корень уравнения: 16^x=64
Решение
Подробное решение
Дано уравнение:1 6 x = 64 16^{x} = 64 1 6 x = 64 или1 6 x − 64 = 0 16^{x} - 64 = 0 1 6 x − 64 = 0 или1 6 x = 64 16^{x} = 64 1 6 x = 64 или1 6 x = 64 16^{x} = 64 1 6 x = 64 - это простейшее показательное ур-ние Сделаем заменуv = 1 6 x v = 16^{x} v = 1 6 x получимv − 64 = 0 v - 64 = 0 v − 64 = 0 илиv − 64 = 0 v - 64 = 0 v − 64 = 0 Переносим свободные слагаемые (без v) из левой части в правую, получим:v = 64 v = 64 v = 64 Получим ответ: v = 64 делаем обратную замену1 6 x = v 16^{x} = v 1 6 x = v илиx = log ( v ) log ( 16 ) x = \frac{\log{\left(v \right)}}{\log{\left(16 \right)}} x = log ( 16 ) log ( v ) Тогда, окончательный ответx 1 = log ( 64 ) log ( 16 ) = 3 2 x_{1} = \frac{\log{\left(64 \right)}}{\log{\left(16 \right)}} = \frac{3}{2} x 1 = log ( 16 ) log ( 64 ) = 2 3
График
-12.5 -10.0 -7.5 -5.0 -2.5 0.0 2.5 5.0 7.5 10.0 12.5 15.0 0 100000000000000
x 1 = 3 2 x_{1} = \frac{3}{2} x 1 = 2 3 log(8) pi*I
x2 = -------- - --------
2*log(2) 2*log(2) x 2 = log ( 8 ) 2 log ( 2 ) − i π 2 log ( 2 ) x_{2} = \frac{\log{\left(8 \right)}}{2 \log{\left(2 \right)}} - \frac{i \pi}{2 \log{\left(2 \right)}} x 2 = 2 log ( 2 ) log ( 8 ) − 2 log ( 2 ) iπ log(8) pi*I
x3 = -------- + --------
2*log(2) 2*log(2) x 3 = log ( 8 ) 2 log ( 2 ) + i π 2 log ( 2 ) x_{3} = \frac{\log{\left(8 \right)}}{2 \log{\left(2 \right)}} + \frac{i \pi}{2 \log{\left(2 \right)}} x 3 = 2 log ( 2 ) log ( 8 ) + 2 log ( 2 ) iπ 3 pi*I
x4 = - + ------
2 log(2) x 4 = 3 2 + i π log ( 2 ) x_{4} = \frac{3}{2} + \frac{i \pi}{\log{\left(2 \right)}} x 4 = 2 3 + log ( 2 ) iπ
Сумма и произведение корней
[src] log(8) pi*I log(8) pi*I 3 pi*I
0 + 3/2 + -------- - -------- + -------- + -------- + - + ------
2*log(2) 2*log(2) 2*log(2) 2*log(2) 2 log(2) ( ( ( 0 + 3 2 ) + ( log ( 8 ) 2 log ( 2 ) − i π 2 log ( 2 ) ) ) + ( log ( 8 ) 2 log ( 2 ) + i π 2 log ( 2 ) ) ) + ( 3 2 + i π log ( 2 ) ) \left(\left(\left(0 + \frac{3}{2}\right) + \left(\frac{\log{\left(8 \right)}}{2 \log{\left(2 \right)}} - \frac{i \pi}{2 \log{\left(2 \right)}}\right)\right) + \left(\frac{\log{\left(8 \right)}}{2 \log{\left(2 \right)}} + \frac{i \pi}{2 \log{\left(2 \right)}}\right)\right) + \left(\frac{3}{2} + \frac{i \pi}{\log{\left(2 \right)}}\right) ( ( ( 0 + 2 3 ) + ( 2 log ( 2 ) log ( 8 ) − 2 log ( 2 ) iπ ) ) + ( 2 log ( 2 ) log ( 8 ) + 2 log ( 2 ) iπ ) ) + ( 2 3 + log ( 2 ) iπ ) log(8) pi*I
3 + ------ + ------
log(2) log(2) log ( 8 ) log ( 2 ) + 3 + i π log ( 2 ) \frac{\log{\left(8 \right)}}{\log{\left(2 \right)}} + 3 + \frac{i \pi}{\log{\left(2 \right)}} log ( 2 ) log ( 8 ) + 3 + log ( 2 ) iπ / log(8) pi*I \ / log(8) pi*I \ /3 pi*I \
1*3/2*|-------- - --------|*|-------- + --------|*|- + ------|
\2*log(2) 2*log(2)/ \2*log(2) 2*log(2)/ \2 log(2)/ 1 ⋅ 3 2 ( log ( 8 ) 2 log ( 2 ) − i π 2 log ( 2 ) ) ( log ( 8 ) 2 log ( 2 ) + i π 2 log ( 2 ) ) ( 3 2 + i π log ( 2 ) ) 1 \cdot \frac{3}{2} \left(\frac{\log{\left(8 \right)}}{2 \log{\left(2 \right)}} - \frac{i \pi}{2 \log{\left(2 \right)}}\right) \left(\frac{\log{\left(8 \right)}}{2 \log{\left(2 \right)}} + \frac{i \pi}{2 \log{\left(2 \right)}}\right) \left(\frac{3}{2} + \frac{i \pi}{\log{\left(2 \right)}}\right) 1 ⋅ 2 3 ( 2 log ( 2 ) log ( 8 ) − 2 log ( 2 ) iπ ) ( 2 log ( 2 ) log ( 8 ) + 2 log ( 2 ) iπ ) ( 2 3 + log ( 2 ) iπ ) 3*(pi*I + log(8))*(-pi*I + log(8))*(2*pi*I + log(8))
----------------------------------------------------
3
16*log (2) 3 ( log ( 8 ) − i π ) ( log ( 8 ) + i π ) ( log ( 8 ) + 2 i π ) 16 log ( 2 ) 3 \frac{3 \left(\log{\left(8 \right)} - i \pi\right) \left(\log{\left(8 \right)} + i \pi\right) \left(\log{\left(8 \right)} + 2 i \pi\right)}{16 \log{\left(2 \right)}^{3}} 16 log ( 2 ) 3 3 ( log ( 8 ) − iπ ) ( log ( 8 ) + iπ ) ( log ( 8 ) + 2 iπ ) x2 = 1.5 - 2.2661800709136*i x3 = 1.5 + 2.2661800709136*i x4 = 1.5 + 4.53236014182719*i