# If lg2, lg(2^x - 1) and lg(2^x + 3 ) are in A.P., calculate the value of x.

*print*Print*list*Cite

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)

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)**

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.