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Super, thanks!!!!
Thanks for watching! 💯🙏🤩💕I'm glad you found it helpful! 🔥🥰✅💕
x1=log(3+V8)/log(log2/log3) , x2=log(3-V8)/log(log2/log3) , test , x1 --> result 6 , test x2 --> result 6 , OK ,
[log_3(2)]ⁿ+[log_2(3)]ⁿ=6Let h=log_3(2) and (1/h)=log_2(3)hⁿ+(1/h)ⁿ=6hⁿ+[1/(hⁿ)]=6Let a=hⁿa+(1/a)=6a²+1=6aa²-6a+1=0a²-6a+9=8(a-3)²=8|a-3|=2√2a-3=±2√2a=3±2√2hⁿ=3±2√2n=log_h(3±2√2)n=log_[log_3(2)](3±2√2) ❤❤
(log₂3)ˣ = uu + 1/u = 6u² - 6u + 1 = 0u = 3 ± 2√2 = (√2 ± 1)²(log₂3)ˣ = (√2 ± 1)²xln(log₂3) = 2ln(√2 ± 1)x = 2ln(√2 ± 1)/ln(log₂3)
Thanks for sharing your method and feedback 💯🙏🤩💕
Super, thanks!!!!
Thanks for watching! 💯🙏🤩💕I'm glad you found it helpful! 🔥🥰✅💕
x1=log(3+V8)/log(log2/log3) , x2=log(3-V8)/log(log2/log3) , test , x1 --> result 6 , test x2 --> result 6 , OK ,
[log_3(2)]ⁿ+[log_2(3)]ⁿ=6
Let h=log_3(2) and (1/h)=log_2(3)
hⁿ+(1/h)ⁿ=6
hⁿ+[1/(hⁿ)]=6
Let a=hⁿ
a+(1/a)=6
a²+1=6a
a²-6a+1=0
a²-6a+9=8
(a-3)²=8
|a-3|=2√2
a-3=±2√2
a=3±2√2
hⁿ=3±2√2
n=log_h(3±2√2)
n=log_[log_3(2)](3±2√2) ❤❤
(log₂3)ˣ = u
u + 1/u = 6
u² - 6u + 1 = 0
u = 3 ± 2√2 = (√2 ± 1)²
(log₂3)ˣ = (√2 ± 1)²
xln(log₂3) = 2ln(√2 ± 1)
x = 2ln(√2 ± 1)/ln(log₂3)
Thanks for sharing your method and feedback 💯🙏🤩💕