Good nigth Dear Sir. Congratulattions for your excellent explanation to solve this question !!!! Please, i would like resquet your help to solve the folowing equation: e^x+x=2 (can not achive solution until now )
By simply looking at the equation and guessing, I saw the following solutions within seconds: x = 1: 2^(1^2-1) = 2^0 = 1 = 1 x = 1/√2: 2^((1/√2)^2-1) = 2^(1/2-1) = 2^(-1/2) = 1/√2 Why does it take 17 minutes to overlook an obvious solution? If you work with the W function, you should know its two arms W0 and W-1.
x =1 is not unique. you lost one solution. you lost one case. 2^(x² -1) =x 2^x² /2 =x 1 /2 =x×2^-x² squaring both sides. 1 /4 =x²×2^-2x² multiplying by -2 both sides. -1 /2 =-2x²×2^-2x² -1 /2 =-2x²×(e^ln2)^-2x² -1 /2 =-2x²×e^(-ln2×2x²) -ln2 /2 =-2x²ln2×e^(-ln2×2x²) ln(1 /2) /1 /2 =-2x²ln2×e^(-ln2×2x²). apply both sides lambert W function. ln(1 /2) =-2x²ln2 -ln2 =-2x²ln2 2x² =1, x =1 /√2. we reject negative solution. verify this solution. 2^(1 /√2)² -1 =1 /√2 2^(1 /2 -1) =1 /√2 2^(-1 /2) =1 /√2. this is second solution. x1 =1 /√2. x2 =1.
i dont think you need Lambert function. starting from -1/2=-2x^2 . 2^(2x^2) 1/2=a.2^- a , where a=2x^2 rearranging 2^a=2a by inspection a=1 and a=2 are solutions. this corresponds to x=1/√2 and x=1
Way of explanation is very good 👌
Thank you so much! ❤
Good nigth
Dear Sir.
Congratulattions for your excellent explanation to solve this question !!!!
Please, i would like resquet your help to solve the folowing equation:
e^x+x=2 (can not achive solution until now )
As soon as possible! ❤
@@SALogics Thanks so much for your attention.
best regards
Seems like a lot of work for x=1, I figured it out in first guess
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2^(xx)=2x square both sides
2^(2xx) =4xx subs a=2xx
2^a=2a
take log base 2
a =1+log_2(a)
a=1 or 2
x=1/√2 or 1
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I solved 80 percent on my own. Merry Christmas, from a math minor an
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s psychology over doing it
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By simply looking at the equation and guessing, I saw the following solutions within seconds:
x = 1: 2^(1^2-1) = 2^0 = 1 = 1
x = 1/√2: 2^((1/√2)^2-1) = 2^(1/2-1) = 2^(-1/2) = 1/√2
Why does it take 17 minutes to overlook an obvious solution? If you work with the W function, you should know its two arms W0 and W-1.
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There are 2 solutions. Check your work by graphing y = 2^(x^2 -1) and y = x. Notice 2 intersections. Therefore, 2 real solutions.
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A Nice Exponential Equation: 2^(x² - 1) = x; x =?
2 > x > 0; 2[2^(x² - 1)] = 2(x), 2^x² = 2x, (2^x²)² = (2x)², (2²)^x² = (2²)x²
4^x² = 4x², (4x²)^(1/x²) = (4^x²)^(1/x²) = 4
(4x²)^(1/x²) = 4 = [4(1)]^(1/1) or (4x²)^(1/x²) = 4 = 2² = [4(1/2)]^[1/(1/2)]
x² = 1, x = 1 > 0 or x² = 1/2, x = 1/√2 > 0
Answer check:
x = 1: 2^(x² - 1) = 2^(1 - 1) = 2⁰ = 1 = x; Confirmed
x = 1/√2: 2^(1/2 - 1) = 2^(- 1/2) = 1/√2 = x; Confirmed
Final answer:
x = 1 or x = 1/√2 = (√2)/2
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BRUH ITS EASY FROM THE THUMBNAIL ITS ONE
2^((1^2 which is 1)-1 which is 0) so that is 2^0
2^0=1
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I think one by root two is also a correct value of x in this equation please justify it
Yes, you are right!❤
x =1 is not unique. you lost one solution. you lost one case.
2^(x² -1) =x
2^x² /2 =x
1 /2 =x×2^-x²
squaring both sides.
1 /4 =x²×2^-2x²
multiplying by -2 both sides.
-1 /2 =-2x²×2^-2x²
-1 /2 =-2x²×(e^ln2)^-2x²
-1 /2 =-2x²×e^(-ln2×2x²)
-ln2 /2 =-2x²ln2×e^(-ln2×2x²)
ln(1 /2) /1 /2 =-2x²ln2×e^(-ln2×2x²).
apply both sides lambert W function.
ln(1 /2) =-2x²ln2
-ln2 =-2x²ln2
2x² =1, x =1 /√2.
we reject negative solution.
verify this solution.
2^(1 /√2)² -1 =1 /√2
2^(1 /2 -1) =1 /√2
2^(-1 /2) =1 /√2.
this is second solution.
x1 =1 /√2.
x2 =1.
Yes correct value of x is one by root two
i dont think you need Lambert function.
starting from -1/2=-2x^2 . 2^(2x^2)
1/2=a.2^- a , where a=2x^2
rearranging
2^a=2a
by inspection a=1 and a=2 are solutions.
this corresponds to x=1/√2 and x=1
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X=1
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