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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hala718's profile pic

Posted on

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.

 

neela's profile pic

Posted on

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.

 

 

 

thewriter's profile pic

Posted on

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.

giorgiana1976's profile pic

Posted on

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!

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