Determine all possible pairs (A, B) of
distinct real numbers such that:
A, A
^{ln(A) }, B
^{ln(B)} and (A*B)
^{ln(A*B)} are consecutive terms of a geometric sequence.
Prove that there are no others.
*** ln(x) denotes the
natural logarithm of x.
I don't know if it is unique, but here is the one solution I found:
If you take the logarithm of terms in geometric sequence the resulting terms are in arithmetic sequence:
ln(A), (ln(A))^2, (ln(B))^2, (ln(AB))^2
with common difference d=(ln(A))^2ln(A)
(ln(B))^2=ln(A) + 2d=2(ln(A))^2  ln(A)
(ln((AB))^2 = ln(A) + 3d
(ln(A)+ln(B))^2 = 3(ln(A))^2  2ln(A)
(ln(A))^2 + 2ln(A)ln(B) + ln(B)^2 = 3(ln(A))^2  2ln(A)
substitute
(ln(A))^2 + 2ln(A)ln(B) + 2(ln(A))^2  ln(A) = 3(ln(A))^2  2ln(A)
2ln(A)ln(B) = ln(A)
ln(B)= 1/2
B= e^(1/2)
back to the original sequence we have common ratios
A^ln(A) / A = B^ln(B) / A^ln(A)
A^(2ln(A)1) = e^(1/4)
taking logarithms yields the quadratic
2(ln(A))^2  ln(A)  1/4 = 0
of the two solutions the one that works is
ln(A)=(1√3)/4
A=e^((1√3)/4)
the four terms have the constant ratio √3/8

Posted by Jer
on 20140221 23:17:21 