If lg2, lg(2^x - 1) and lg(2^x + 3 ) are in A.P., calculate the value of x.
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log 2 , log(2^x - 1), log(2^x + 3 ) ..... find f
Since the above is an arthimatical progression, then:
log (2^x - 1) - log2 = log (2^x + 3) - log (2^x -1)
But we know that log a - log b = log a/b
==> log (2^x -1)/2 = log (2^x+3)/(2^x -1)
==> (2^x -1)/2 = (2^x + 3)/(2^x -1)
Cross multiply:
==? (2^x -1)^2 = 2(2^x +3)
Expand brackets:
==> 2^2x -2*2^x + 1 = 2*2^x + 6
==> (2^x)^2 - 4*2^x -5 =0
Let y = 2^x
==> y^2 - 4y -5 = 0
==> (y-5)(y+1) = 0
==> y= 5 ==> 5= 2^x ==> x = log 2 (5)
==? y= -1 ==> -1 = 2^x (impossible)
Then the solution is :
x = log 2 (5)
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Since the terms are in AP, the diffrence between consecutive terms should be same.
2nd term -1st term = 3rd term - 2nd term
lg(2^x -1) - lg2 = lg(2^x+3)- lg(2^x-1).
lg(2^x-1)/2} = lg (2^x+3)/2^x -1) as lga- lgb =lg(a/b). Take antilog.
(2^x-1)/2 = (2^x+3)/2^x-1)
(t-1)/2 = (t+3)/(t-1). Here t = 2^x.
(t-1)^2 = 2(t+3)
t^2-2t+1 = 2t+6
t^2-4t-5 = 0
(t-5)(t+1) = 0
t-5 = 0 or t+1 = 0
t-5 = 0 gives t = 5 Or 2^x = 5 Or x = log 2(5).
t+1 = 0 gives = t =-1 , Or 2^x = -1 which has no real solution.
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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