Time (8:20) You need to work with LN function instead of LOG function b must be in this interval: 1 < b < 2 Using LN function W(32 Ln 2) / Ln 2 = 2.27655823 / 0.693147181 = 3.28437927 b = 5 - 3.28437927 = 1.715620733 Right solution Using LOG function W(32 log 2) / log 2 = 1.72181283 / 0.301029996 = 5.719738415 b = 5 - 5.719738415 = --0.719738415 Wrong solution
2^1 + 1 = 3 < 5 and 2^2 + 2 = 6 > 5 and 2^x + x has an always positive derivative so it's a monotonically increasing function. Since 3 < 5 < 6, therefore 1 < b < 2. For such simply differentiable and continuous function based equations, no unusual (like Lambert W) function is needed and a solution to any accuracy is easily numerically computable. e.g. here b = 1.715620733275586169380916428210115405349201542402693776216135036789959346078769637168046686169244547... (100 places of decimal)
Hello Math Master. There is an error (time 3:33) "2 = e ** log 2" You should write "2 = e ** Ln 2" You need to work with LN function, not with LOG function e ** log 2 = 1.351249873 e ** Ln 2 = 2
2+1=3, 4+2=6. Значит, а где-то между 1 и 2. Обозначим b как 2-х. Тогда 4/2^x+2-x=5⇔4/2^x-х=3. 4/2^x, в свою очередь, будет между 2 и 1 так же, т. е. эти слагаемые в одном интервале. Максимально 1-2, минимально 2-1... Похоже, действительных корней нет...
How and where did you evaluate the W function for log(2)^32? I keep hearing about wolfram alpha or other sophisticated software to evaluate the W function. How does that help anybody solve the problem? Am I supposed to just know the answer to the W function for log(2)^32?
b=5. -- w(32 Ln 2)/Ln 2
= 5 -- ( ~2, 27) /Ln 2)
= 5 --3 ,27= (~1/7)
2^1,7 + 1,7=. (~5)
🙏
Time (8:20) You need to work with LN function instead of LOG function
b must be in this interval: 1 < b < 2
Using LN function
W(32 Ln 2) / Ln 2 = 2.27655823 / 0.693147181 = 3.28437927
b = 5 - 3.28437927 = 1.715620733
Right solution
Using LOG function
W(32 log 2) / log 2 = 1.72181283 / 0.301029996 = 5.719738415
b = 5 - 5.719738415 = --0.719738415
Wrong solution
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2^1 + 1 = 3 < 5 and 2^2 + 2 = 6 > 5 and 2^x + x has an always positive derivative so it's a monotonically increasing function. Since 3 < 5 < 6, therefore 1 < b < 2.
For such simply differentiable and continuous function based equations, no unusual (like Lambert W) function is needed and a solution to any accuracy is easily numerically computable.
e.g. here b = 1.715620733275586169380916428210115405349201542402693776216135036789959346078769637168046686169244547... (100 places of decimal)
Hello Math Master. There is an error (time 3:33) "2 = e ** log 2" You should write "2 = e ** Ln 2"
You need to work with LN function, not with LOG function
e ** log 2 = 1.351249873
e ** Ln 2 = 2
2+1=3, 4+2=6. Значит, а где-то между 1 и 2. Обозначим b как 2-х. Тогда 4/2^x+2-x=5⇔4/2^x-х=3. 4/2^x, в свою очередь, будет между 2 и 1 так же, т. е. эти слагаемые в одном интервале. Максимально 1-2, минимально 2-1... Похоже, действительных корней нет...
How and where did you evaluate the W function for log(2)^32? I keep hearing about wolfram alpha or other sophisticated software to evaluate the W function. How does that help anybody solve the problem? Am I supposed to just know the answer to the W function for log(2)^32?
wolfram alpha
Please tell me
e the title of the background music.