# Solve for x if ln(2^x - 1), ln(2^x + 3), ln(2^x + 5) are the terms of an a.s.Solve for x if ln(2^x - 1), ln(2^x + 3), ln(2^x + 5) are the terms of an a.s.

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First, we'll note the terms of the sequence:

a1 = ln(2^x - 1)

a2 = ln(2^x + 3 )

a3 = ln(2^x + 5 )

If the given sequence is an a.p., then:

a2 - a1 = a3 - a2

ln(2^x + 3 ) - ln(2^x - 1) = ln(2^x + 5 ) - ln(2^x + 3 )

Because the bases of logarithms are matching, we'll use the quotient property of the logarithms:

ln [(2^x + 3 )/(2^x - 1)] = ln [(2^x + 5 )/(2^x + 3 )]

We'll use the one to one property:

[(2^x + 3 )/(2^x - 1)] = [(2^x + 5 )/(2^x + 3 )]

We'll remove the brackets and cross multiply:

(2^x - 1)*(2^x + 5 ) = (2^x + 3 )^2

We'll remove the brackets and we'll expand the square from the right side:

2^2x +4*2^x - 5 = 2^2x + 6*2^x + 9

We'll eliminate like terms:

-2*2^2x - 14 = 0

We'll divide by -2:

2^2x + 7 = 0

2^2x = -7 impossible!

**The equation has no solutions!**

I assume that you mean arithmetic progression by a.s.

As ln(2^x-1), ln(2^x+3) and ln(2^x+5) are in A.P.

2*ln(2^x+3)= ln(2^x+5)+ln(2^x-1)

We know that ln a+ ln b=ln(a*b) and p ln x= ln(x^p)

Therefore: ln[(2^x+3)^2]= ln[(2^x+5)*(2^x-1)]

Take the antilog on both the sides

(2^x+3)^2= (2^x+5)*(2^x-1)

Let 2^x=t

we have:

(t+3)^2= (t+5)*(t-1)

=>t^2+6t+9=t^2+4t-5

=>2t=-14

=>t=-7

As t=2^x, 2^x=-7. This is never possible as 2 is positive.

Therefore the equation has no solution.

Since th terms of arithmetic series should have the same (constant) common difference beteen the succesive terms, we ghave:

2nd term - 1st term = 3rd term - 2nd term:

ln(2^x+3)- ln(2^x-1) = ln(2^x+5)- ln(2^x+3)

Using the property of logarithms , ln a- lnb = ln(a/b) we get:

ln {(2^x-3)/(2^x -1) }= ln {(2^x+5)/(2^x+3)}. Take antilogarithms:

(2^x+3)/(2^x-1) = (2^x+5)/(2^x+3). Multiply by LCM denominators, (2^x+3)(2^x-1) both sides.

(2^x+3)^2 = (2^x+5)(2^x-1). Put 2^x =t.

(t+3)^2 = (t+5)(t-1)

t^2+6t+9 = t^2+4t-5

2t+9 = -5

2t = -5-9 = -14

t = -14/2 =-7.

So 2^x = -7. Take logarithms:

x ln 2 = -7 = 7 cos(pi+isinpi)

x = (7/ln2) e^ (i*(pi)) . The solution is a complex number.

ln (2^x-1), ln (2^x + 3) , ln (2^x + 5)

==> ln (2^x +3) - ln (2^x -1) = ln (2^x + 5) - ln (2^x +3)

But we know that:

ln a - ln b = ln a/b

==> ln (2^x +3)/(2^x -1) = ln (2^x +5)/(2^x +3)

==> (2^x +3)/(2^x-1) = (2^x +5)/(2^x +3)

Cross multiply:

==> (2^x + 3)^2 = (2^x -1)(2^x + 5)

==> 2^2x + 6*2^x + 9 = 2^2x + 4*2^x - 5

Now group similar:

==> 2*2^x = -14

==> 2^x = -7

This is impossible because 2^x is positive and -7 is negative.

Then there is no solution for the problem.