Find, if possible, Constant k

Is there a constant k such that the equation e^(x) = 2^(kx) holds true for all values of x?

e^x = 2^(kx)
e^x = (2^k)^x
e = 2^k
Take logarithms.
ln(e) = ln(2^k)
1 = kln(2)
k = 1/ln(2)

rearrange

e^x = 2^(kx)
ln e^x = ln (2^(kx))

x= kx ln 2
k ln 2 = 1
k = 1/ln(2)

lets check with random numbers
x=1
e^x = 2.71
2^1/ln2 = 2.71

x = 14.5
e^14.5 = 1 982 759.26
2^14.5/ln2 = 1 982 759.26

looks good to me, any q's?

You have to solve for k

taking ln on both sides

ln(e^x) = kx ln(2)

x = kx ln(2)

1 = k ln(2)

k = 1/ln 2

Yes. If k = 1/ln2 then 2^(kx) = 2^(x/ln2) = e^x.

k = 1.4427 you ask a lotta questions today