Question 1011381
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Find all values of k such that equation 2^x-k(2^-x)=1 has {{{highlight(cross(exact))}}} exactly one real solution. 
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{{{2^x-k*2^(-x)}}} = {{{1}}}.      (1)

Introduce new variable y = {{{2^x}}}. Then the equation takes the form

y - k*{{{(1/y)}}} = {{{1}}}.

Multiply both sides by y. You will get

{{{y^2 - k}}} = {{{y}}},    or

{{{y^2 - y - k}}} = {{{0}}}.         (2)

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

d = 0,

where d = {{{sqrt(b^2 - 4ac)}}} = {{{sqrt((-1)^2 + 4k)}}} = {{{sqrt(1 + 4k)}}}.

It means that 1 + 4k = 0,   or   k = {{{-1/4}}}.

Then the equation (2) become 

{{{y^2 - y + 1/4}}} = {{{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 = {{{1/2}}}.

Thus the equation (1) has the unique root if and only if k = {{{-1/4}}}. 

This root is x = -1.
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