Harvard University Entrance Exam || Logarithms Problem Tricks || 99% Failed Admission Aptitude Test

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Harvard University Exam Question || Algebra Problem || Aptitude Simplification Test
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Пікірлер: 19
@baselinesweb
@baselinesweb 20 күн бұрын
Seems like polynomial division would be a much easier way to go. Where is the backup for your statistic that 99% failed? That isn't believable.
@TommyRaines
@TommyRaines 20 күн бұрын
Its clickbait
@ichdu6710
@ichdu6710 19 күн бұрын
@@TommyRaines which dont work anymore
@michaelbialas4026
@michaelbialas4026 17 күн бұрын
Mit Hilfe der Polynomdivision hatte ich die Lösung in 2 Minuten gefunden.
@HabibAkili
@HabibAkili 20 күн бұрын
8^x + 4^x = 36 36 - 4^x = 8^x 6² - 2²^x = 6² - (2^x)² ( 6-2^x) ( 6+2^x) = 8^x = ( 2*4)^x = 2^x * 4^x 6+2^x= 4^× since 6-2^x < 6+2^x 4^x - 2^x = 6 2^x ( 2^x-1)=6 6-2^x= 2^x 2(2^x) =6 2^x=6/2=3 x log 2 = log 3 x = log 3/log2...
@davidtaran952
@davidtaran952 16 күн бұрын
y=2^x>0 y^3+y^2-36=0 (1) Of the factors 36, the number y=3 is the root: 2^x=3 -> x1=log2(3). Let's divide equation (1) by the binomial (y-3). We get: y^2+4y+12=0 with a negative determinant, i.e. complex roots.
@RyanLewis-Johnson-wq6xs
@RyanLewis-Johnson-wq6xs 20 күн бұрын
8^(Log[2,3])+4^(Log[2,3])=36 x=Log[2,3]
@MARTINWERDER
@MARTINWERDER 20 күн бұрын
Transform the equation in to 2^3^X + 2^2^X = 36 Substitute 2^X = a, new equation is a^3 + a^2 = 36 = a^2 (a + 1) = 36 a^2 is much greater than (a + 1). Which factors of 36 have an integer square root and show a significant difference in size? It's the pair 9 / 4 , and 9 is a^2 a=3 = 2^X, and X = 3 log base 2 = 1,58496 8^X + 4^X = 27 + 9 = 36
@ichdu6710
@ichdu6710 19 күн бұрын
other factorizations must be considered as nothing else is mentioned here
@RyanLewis-Johnson-wq6xs
@RyanLewis-Johnson-wq6xs 20 күн бұрын
It’s in my head.
@RealQinnMalloryu4
@RealQinnMalloryu4 20 күн бұрын
36/8^x=4.4Log^c36/4=9 {4.4+9}=5.3 (x ➖ 5x+3).
@prollysine
@prollysine 19 күн бұрын
by fakt. , let u=2^x , add -12u , 12u , u^3-3u^2 +4u^2-12u +12u-36=0 , u^2(u-3)+4u(4-3)+12(u-3)=0 , (u-3)(u^2+4u+12)=0 , u-3=0 , u=3 , recall , u=2^x , 2^x=3 , x=log3/log2 , test , 8^(log3/log2)+4^(log3/log2)=27+9 , 27+9=36 , same , OK , / for complex , u^2+4u+12=0 , /
@TommyRaines
@TommyRaines 20 күн бұрын
The cube root of 4 is a good approximation
@tunneloflight
@tunneloflight 20 күн бұрын
x = lg2(3)
@ahmedmahrous7787
@ahmedmahrous7787 14 күн бұрын
Very complicated answer
@superacademy247
@superacademy247 14 күн бұрын
It's not 😂. Go through it over and over again. It's pretty straight forward.
@簡欽慧
@簡欽慧 18 күн бұрын
太複雜了
@superacademy247
@superacademy247 17 күн бұрын
It's straight forward!
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