If you posit right from the start that x=2ᵃ, the initial equation can immediately be rewritten as x + 1/x = 4, and then as x² - 4x + 1 = 0. This feels more natural (at least to me) than multiplying both sides by 2ᵃ and using the substitution afterwards. In any event, nice little problem, very well explained.
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My way, a = 1,9 1) 2^a + 1/2^a = 4 2) trick, 2^a = Y 3) Y² - 4Y + 1 = zero By baskara ou Google, Y' = 3,7321 ; Y" = 0,2679 ( I will use the first, but...) 4) 2^a = 3,7321 a = 1,9 ( using my cell calculater ) Prove: 3,7321 + 1 / 3,7321 = 4 4 = 4 Bingo from Brazil !!!!!!
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If you posit right from the start that x=2ᵃ, the initial equation can immediately be rewritten as x + 1/x = 4, and then as x² - 4x + 1 = 0. This feels more natural (at least to me) than multiplying both sides by 2ᵃ and using the substitution afterwards. In any event, nice little problem, very well explained.
nice dear
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This can be rewritten as cosh(a*ln(2)) = 2 and so a = +-arcosh (2)/ln(2)
On [1 , oo[ arcosh is ln(x + sqrt (x^2 - 1))
So a = +-ln(2 + sqrt(3))/ln(2)
nice job, thanks
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^=read as to the power
*=read as square root
a={log(2+*3)}/log2
Or
a= {log(2-*3)/log2......May be
okay dear
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My way, a = 1,9
1) 2^a + 1/2^a = 4
2) trick, 2^a = Y
3) Y² - 4Y + 1 = zero
By baskara ou Google, Y' = 3,7321 ; Y" = 0,2679 ( I will use the first, but...)
4) 2^a = 3,7321
a = 1,9 ( using my cell calculater )
Prove: 3,7321 + 1 / 3,7321 = 4
4 = 4
Bingo from Brazil !!!!!!
beautiful way
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Didn't see the video. No paper
Result is ln(2+√3)/ln(2).
Replace 2^a=x; x +1/x=4
x^2 +1 = 4x ; x^2 - 4x -1 =0
Complete squares
(x+2)^2 -3 = 0
x= 2 + √3
a= ln (2+√3)/ln(2)
thanks for solution
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Too many innecessary steps 😢
only explanation to clear concepts
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