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 Naturally Geometric (Posted on 2014-02-21)
Determine all possible pairs (A, B) of distinct real numbers such that:
A, Aln(A) , Bln(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.

 No Solution Yet Submitted by K Sengupta No Rating

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 A solution Comment 1 of 1
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))^2-ln(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 2014-02-21 23:17:21
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