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Solve for x : x^(1+log 2 sqrtx) = 16

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realcomplexnr | Student, Grade 10 | eNoter

Posted August 20, 2010 at 12:54 AM via web

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Solve for x : x^(1+log 2 sqrtx) = 16

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giorgiana1976 | College Teacher | Valedictorian

Posted August 20, 2010 at 1:00 AM (Answer #1)

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This is a logarithmic equation.

We'll take logarithms both side of the equation and the logarithms will have the base 2:

log 2 [x^(1+log 2 sqrtx)] = log 2 16

We'll use the power property of logarithms and we'll write 16 = 2^4:

(1+log 2 sqrtx) * log 2 x= log 2 2^4

[1+log 2 (x)^1/2] * log 2 x= 4*log 2 2

{1 + [log 2 (x)]/2}* log 2 x= 4

We'll remove the brackets form the left side:

log 2 x + [(log 2 x)^2]/2 - 4 = 0

2*log 2 x + (log 2 x)^2 - 4 = 0

We'll substitute log 2 x = t

t^2 + 2t - 4 = 0

We'll apply the quadratic formula:

t1 = [-2+sqrt(4+16)]/2

t1 = (-2+2sqrt5)/2

t1 = -1+sqrt5

t2 = -1-sqrt5

 log 2 x = t1

x1 = 2^-1+sqrt5

x2 = 2^-(1+sqrt5)

x2 = 1/2^(1+sqrt5)

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted August 20, 2010 at 1:28 AM (Answer #2)

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x^(1+log 2 sqrtx) = 16

x^(1+log 2 (sqrtx) = 2^4

==> log 2 [x^(1+log 2 (sqrtx)] = log 2 2^4

==> [1+log 2 (sqrtx)]*log 2 x] = 4*log 2 2 = 4*1= 4

==> [1+ (1/2)log 2 x] log 2 x = 4

==> log 2 x + (1/2)(log 2 x)^2 = 4

==> (log 2 x)^2 + 2 log 2 x - 8 = 0

==> (log 2 x +4)(log 2 x - 2) = 0

==> log 2 x = -4

==> x1= 2^-4 = 1/16

==> x1= 1/16

==> log 2 x = 2

==> x2= 4

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jeyaram | Student , Undergraduate | Valedictorian

Posted August 20, 2010 at 9:16 PM (Answer #3)

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x^(1+log 2 sqrtx) = 16

log 2 [x^(1+log 2 sqrtx)] = log 2 16

log 2 x+1/2log 2 x* log 2 x=4

put   log 2 x =j

j+1/2j^2=4

j^2 + 2j -8=0

(j+4)(j-2)=0

j=(-4)          or           j=2

log 2 x=(-4)         or       log 2 x=2

x=0.0625         or     x=4

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