Question

For what value of k does the equation e^(2x)=k*sqrt(x) have exactly one solution?

Answer 1

wed have to scale sqrt(x) to reach i spose

Answer 2

somehow

Answer 3

e^(2x)=k*sqrt(x)

2x = ln(k*sqrt(x))

2x = ln(k) + ln(sqrt(x))

2x-ln(sqrt(x)) = ln(k)

maybe in the right direction?

Answer 4

im nt so sure but that wht i have
k has to be bigger than zero, because exp(2x)>0 en sqrt(x) > 0.
Second : x >= 0, because sqrt(x) only defined for real values x >= 0
For x >= 0, exp(2x) >= 1, so k sqrt(x) >= 1
Take x = 0, then we have 1 = k * 0, so k has to be +infinity
Take x = 0.5 then we have e = k / sqrt(2) => k = e sqrt(2)
Take x = 1, then we have e² = k, so k has to be e²
Take x = 2, then we have e^4 = k sqrt(2), so k has to be e^4 / sqrt(2)
So we have a minimum for k. We have to calculate that minimum
k = exp(2x) / sqrt(x)
=> k ' = ( 2 exp(2x) sqrt(x) - exp(2x) / 2 sqrt(x) ) / x = 0
2 sqrt(x) - 1/2 sqrt(x) = 0 => 4 x - 1 = 0 => x = 1/4
if x = 1/4, we have exp(0.5) = k / 2 => k = 2 exp(0.5)

So we have exactly one solution for k = 2 exp(0.5) and
zero solutions for k < 2 exp(0.5) and
two solutions for k > 2 exp(0.5)

Answer 5

e^(2x-ln(sqrt(x))) = k

e^(2x) * e^(-ln(sqrt(x))) = k

Answer 6

im nt so sure tho

Answer 7

e^(2x)-k*sqrt(x)=0

i cant be sure either but,

i know its between sqrt(10.8) and sqrt(10.9) :)

Answer 8

i think the wolf says your right http://www.wolframalpha.com/input/?i=e^%282x%29-%282+exp%280.5%29%29*sqrt%28x%29%3D0