For the beginning, let's note the terms of the
sequence:
a1 = lg2
a2 = lg(2^x
- 1)
a3 = lg(2^x + 3 )
If the
given sequence is an a.p., then:
a2 - a1 = a3 -
a2
lg(2^x - 1) - lg2 = lg(2^x + 3 ) - lg(2^x -
1)
We'll use the quotient property of the
logarithms:
lg [(2^x - 1)/2] = lg [(2^x + 3 )/(2^x -
1)]
Because the bases of logarithms are matching, we'll use
the one to one property:
[(2^x - 1)/2] = [(2^x + 3 )/(2^x -
1)]
We'll remove the brackets and cross
multiply:
(2^x - 1)^2 = 2*(2^x + 3
)
We'll expand the square from the left
side:
2^2x - 2*2^x + 1 = 2*2^x +
6
We'll move all terms to one
side:
2^2x - 4*2^x - 5 =
0
We'll substitute 2^x = t
t^2
- 4t - 5 =0
We'll apply the quadratic
formula:
t1 =
[4+sqrt(16+20)]/2
t1 =
(4+6)/2
t1 = 5
t2 =
(4-6)/2
t2 = -1
Since
2^x>0, the only admissible solution is:
2^x =
5
x = log 2
(5)
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