SOLUTION: Find all values of K such that equation 2^x-k(2^-x)=1 has exact one real solution

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Question 1011381: Find all values of K such that equation 2^x-k(2^-x)=1 has exact one real solution
Answer by ikleyn(52787) About Me  (Show Source):
You can put this solution on YOUR website!
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Find all values of k such that equation 2^x-k(2^-x)=1 has highlight%28cross%28exact%29%29 exactly one real solution.
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2%5Ex-k%2A2%5E%28-x%29 = 1.      (1)

Introduce new variable y = 2%5Ex. Then the equation takes the form

y - k*%281%2Fy%29 = 1.

Multiply both sides by y. You will get

y%5E2+-+k = y,    or

y%5E2+-+y+-+k = 0.         (2)

The condition that this equation has only one root is vanishing the discriminant, i.e. 

d = 0,

where d = sqrt%28b%5E2+-+4ac%29 = sqrt%28%28-1%29%5E2+%2B+4k%29 = sqrt%281+%2B+4k%29.

It means that 1 + 4k = 0,   or   k = -1%2F4.

Then the equation (2) become 

y%5E2+-+y+%2B+1%2F4 = 0.         (3)

The discriminant of this equation is zero due to selection of k. (You can check it).

The unique root of this equation is y = 1%2F2.

Thus the equation (1) has the unique root if and only if k = -1%2F4. 

This root is x = -1.