Solving a 'Stanford' University entrance exam | a=?

[@fadetoblah2883]

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.

[@AsadInternationalAcademy]

nice dear

[@AsadInternationalAcademy]

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[@xavierplatiau4635]

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)

[@AsadInternationalAcademy]

nice job, thanks

[@AsadInternationalAcademy]

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[@ManojkantSamal]

^=read as to the power *=read as square root a={log(2+*3)}/log2 Or a= {log(2-*3)/log2......May be

[@AsadInternationalAcademy]

okay dear

[@AsadInternationalAcademy]

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[@JPTaquari]

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 !!!!!!

[@AsadInternationalAcademy]

beautiful way

[@AsadInternationalAcademy]

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[@jacksdu13]

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)

[@AsadInternationalAcademy]

thanks for solution

[@AsadInternationalAcademy]

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[@firebuble]

Too many innecessary steps 😢

[@AsadInternationalAcademy]

only explanation to clear concepts

[@AsadInternationalAcademy]

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