8^x-2^x=60 (уравнение) Учитель очень удивится увидев твоё верное решение 😼
Найду корень уравнения: 8^x-2^x=60
Решение
Подробное решение
Дано уравнение:− 2 x + 8 x = 60 - 2^{x} + 8^{x} = 60 − 2 x + 8 x = 60 или( − 2 x + 8 x ) − 60 = 0 \left(- 2^{x} + 8^{x}\right) - 60 = 0 ( − 2 x + 8 x ) − 60 = 0 Сделаем заменуv = 2 x v = 2^{x} v = 2 x получимv 3 − v − 60 = 0 v^{3} - v - 60 = 0 v 3 − v − 60 = 0 илиv 3 − v − 60 = 0 v^{3} - v - 60 = 0 v 3 − v − 60 = 0 делаем обратную замену2 x = v 2^{x} = v 2 x = v илиx = log ( v ) log ( 2 ) x = \frac{\log{\left(v \right)}}{\log{\left(2 \right)}} x = log ( 2 ) log ( v ) Тогда, окончательный ответx 1 = log ( 2 ) log ( 2 ) = 1 x_{1} = \frac{\log{\left(2 \right)}}{\log{\left(2 \right)}} = 1 x 1 = log ( 2 ) log ( 2 ) = 1 x 2 = log ( log ( − 2 − 11 i ) log ( 2 ) ) log ( 2 ) = log ( log ( − 2 − 11 i ) log ( 2 ) ) log ( 2 ) x_{2} = \frac{\log{\left(\frac{\log{\left(-2 - \sqrt{11} i \right)}}{\log{\left(2 \right)}} \right)}}{\log{\left(2 \right)}} = \frac{\log{\left(\frac{\log{\left(-2 - \sqrt{11} i \right)}}{\log{\left(2 \right)}} \right)}}{\log{\left(2 \right)}} x 2 = log ( 2 ) log ( l o g ( 2 ) l o g ( − 2 − 11 i ) ) = log ( 2 ) log ( l o g ( 2 ) l o g ( − 2 − 11 i ) ) x 3 = log ( log ( − 2 + 11 i ) log ( 2 ) ) log ( 2 ) = log ( log ( − 2 + 11 i ) log ( 2 ) ) log ( 2 ) x_{3} = \frac{\log{\left(\frac{\log{\left(-2 + \sqrt{11} i \right)}}{\log{\left(2 \right)}} \right)}}{\log{\left(2 \right)}} = \frac{\log{\left(\frac{\log{\left(-2 + \sqrt{11} i \right)}}{\log{\left(2 \right)}} \right)}}{\log{\left(2 \right)}} x 3 = log ( 2 ) log ( l o g ( 2 ) l o g ( − 2 + 11 i ) ) = log ( 2 ) log ( l o g ( 2 ) l o g ( − 2 + 11 i ) )
График
-10.0 -7.5 -5.0 -2.5 0.0 2.5 5.0 7.5 10.0 12.5 15.0 17.5 -100000000000 100000000000
/ / ____\\
| |\/ 11 ||
/ ____\ I*|-pi + atan|------||
log\\/ 15 / \ \ 2 //
x2 = ----------- + ----------------------
log(2) log(2) x 2 = log ( 15 ) log ( 2 ) + i ( − π + atan ( 11 2 ) ) log ( 2 ) x_{2} = \frac{\log{\left(\sqrt{15} \right)}}{\log{\left(2 \right)}} + \frac{i \left(- \pi + \operatorname{atan}{\left(\frac{\sqrt{11}}{2} \right)}\right)}{\log{\left(2 \right)}} x 2 = log ( 2 ) log ( 15 ) + log ( 2 ) i ( − π + atan ( 2 11 ) ) / / ____\\
| |\/ 11 ||
/ ____\ I*|pi - atan|------||
log\\/ 15 / \ \ 2 //
x3 = ----------- + ---------------------
log(2) log(2) x 3 = log ( 15 ) log ( 2 ) + i ( π − atan ( 11 2 ) ) log ( 2 ) x_{3} = \frac{\log{\left(\sqrt{15} \right)}}{\log{\left(2 \right)}} + \frac{i \left(\pi - \operatorname{atan}{\left(\frac{\sqrt{11}}{2} \right)}\right)}{\log{\left(2 \right)}} x 3 = log ( 2 ) log ( 15 ) + log ( 2 ) i ( π − atan ( 2 11 ) )
Сумма и произведение корней
[src] / / ____\\ / / ____\\
| |\/ 11 || | |\/ 11 ||
/ ____\ I*|-pi + atan|------|| / ____\ I*|pi - atan|------||
log\\/ 15 / \ \ 2 // log\\/ 15 / \ \ 2 //
2 + ----------- + ---------------------- + ----------- + ---------------------
log(2) log(2) log(2) log(2) ( 2 + ( log ( 15 ) log ( 2 ) + i ( − π + atan ( 11 2 ) ) log ( 2 ) ) ) + ( log ( 15 ) log ( 2 ) + i ( π − atan ( 11 2 ) ) log ( 2 ) ) \left(2 + \left(\frac{\log{\left(\sqrt{15} \right)}}{\log{\left(2 \right)}} + \frac{i \left(- \pi + \operatorname{atan}{\left(\frac{\sqrt{11}}{2} \right)}\right)}{\log{\left(2 \right)}}\right)\right) + \left(\frac{\log{\left(\sqrt{15} \right)}}{\log{\left(2 \right)}} + \frac{i \left(\pi - \operatorname{atan}{\left(\frac{\sqrt{11}}{2} \right)}\right)}{\log{\left(2 \right)}}\right) 2 + log ( 2 ) log ( 15 ) + log ( 2 ) i ( − π + atan ( 2 11 ) ) + log ( 2 ) log ( 15 ) + log ( 2 ) i ( π − atan ( 2 11 ) ) / / ____\\ / / ____\\
| |\/ 11 || | |\/ 11 ||
/ ____\ I*|pi - atan|------|| I*|-pi + atan|------||
2*log\\/ 15 / \ \ 2 // \ \ 2 //
2 + ------------- + --------------------- + ----------------------
log(2) log(2) log(2) 2 + 2 log ( 15 ) log ( 2 ) + i ( − π + atan ( 11 2 ) ) log ( 2 ) + i ( π − atan ( 11 2 ) ) log ( 2 ) 2 + \frac{2 \log{\left(\sqrt{15} \right)}}{\log{\left(2 \right)}} + \frac{i \left(- \pi + \operatorname{atan}{\left(\frac{\sqrt{11}}{2} \right)}\right)}{\log{\left(2 \right)}} + \frac{i \left(\pi - \operatorname{atan}{\left(\frac{\sqrt{11}}{2} \right)}\right)}{\log{\left(2 \right)}} 2 + log ( 2 ) 2 log ( 15 ) + log ( 2 ) i ( − π + atan ( 2 11 ) ) + log ( 2 ) i ( π − atan ( 2 11 ) ) / / / ____\\\ / / / ____\\\
| | |\/ 11 ||| | | |\/ 11 |||
| / ____\ I*|-pi + atan|------||| | / ____\ I*|pi - atan|------|||
|log\\/ 15 / \ \ 2 //| |log\\/ 15 / \ \ 2 //|
2*|----------- + ----------------------|*|----------- + ---------------------|
\ log(2) log(2) / \ log(2) log(2) / 2 ( log ( 15 ) log ( 2 ) + i ( − π + atan ( 11 2 ) ) log ( 2 ) ) ( log ( 15 ) log ( 2 ) + i ( π − atan ( 11 2 ) ) log ( 2 ) ) 2 \left(\frac{\log{\left(\sqrt{15} \right)}}{\log{\left(2 \right)}} + \frac{i \left(- \pi + \operatorname{atan}{\left(\frac{\sqrt{11}}{2} \right)}\right)}{\log{\left(2 \right)}}\right) \left(\frac{\log{\left(\sqrt{15} \right)}}{\log{\left(2 \right)}} + \frac{i \left(\pi - \operatorname{atan}{\left(\frac{\sqrt{11}}{2} \right)}\right)}{\log{\left(2 \right)}}\right) 2 log ( 2 ) log ( 15 ) + log ( 2 ) i ( − π + atan ( 2 11 ) ) log ( 2 ) log ( 15 ) + log ( 2 ) i ( π − atan ( 2 11 ) ) / ____\ / ____\
2 2 2|\/ 11 | |\/ 11 |
log (15) + 4*pi + 4*atan |------| - 8*pi*atan|------|
\ 2 / \ 2 /
------------------------------------------------------
2
2*log (2) − 8 π atan ( 11 2 ) + 4 atan 2 ( 11 2 ) + log ( 15 ) 2 + 4 π 2 2 log ( 2 ) 2 \frac{- 8 \pi \operatorname{atan}{\left(\frac{\sqrt{11}}{2} \right)} + 4 \operatorname{atan}^{2}{\left(\frac{\sqrt{11}}{2} \right)} + \log{\left(15 \right)}^{2} + 4 \pi^{2}}{2 \log{\left(2 \right)}^{2}} 2 log ( 2 ) 2 − 8 π atan ( 2 11 ) + 4 atan 2 ( 2 11 ) + log ( 15 ) 2 + 4 π 2 x2 = 1.95344529780426 - 3.04904281272161*i x3 = 1.95344529780426 + 3.04904281272161*i