The terms 3^x - 1 , 3^(2x + 2), 5*3^x + 1 are the terms of an a.p. Determine the value of x .

4 Answers | Add Yours

hala718's profile pic

hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted on

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

Since the above are terms in an A.P, then:

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

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

Group Similar:

2*3^(2x+2)) - 6*3^x = 0

Divide by 2:

==> 3^(2x+2) - 3*3^x = 0

==> 3^(2x+2) = 3*3^x

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

==> 2x + 2 = x+1

==> x= -1

Top Answer

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll have to determine x, so that the given terms to be the consecutive terms of an arithmetic series.

Now, we'll apply the rule of the consecutive terms of an arithmetical progression. According to the rule, the middle term is the arithmetical mean of the joined terms.

3^(2x + 2) = ( 3^x - 1 + 5*3^x +1)/2

Eliminating like terms, from the brackets, from the right side, we'll get:

3^(2x + 2) = ( 3^x + 5*3^x)/2

Now, we'll factorize by 3^x, to the right side:

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

3^(2x + 2) = 3*3^x

3^(2x + 2) = 3^(x+1)

Because the bases are matching, we'll apply the one to one property:

2x + 2 = x + 1

We'll isolate x to the left side:

2x - x = 1 - 2

x = -1

So, the consecutive terms of the A.P. are:

1/3 - 1 , 3^0, 5/3 + 1

-2/3 , 1 , 8/3

william1941's profile pic

william1941 | College Teacher | (Level 3) Valedictorian

Posted on

As 3^x-1, 3^(2x+2) and 5*3^x+1 are the consecutive terms of an arithmetic progression, we know that:

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

=>3^x+5*3^x=2*3^(2x+2)

=>3^x+5*3^x=2*3^2x*3^2

=>3^x+5*3^x=18*3^2x

=Let 3^x=y, now we have

y+5y=18y^2

=>6y=18y^2

=>y=3y^2

=>3y^2-y=0

=>y(3y-1)=0

=> y=0 or y=1/3

Now as 3^x=y

3^x=0 or 3^x=1/3

3^x=0 is not a valid solution. For 3^x=1/3, x=-1

Therefore the valid value of x is -1.

neela's profile pic

neela | High School Teacher | (Level 3) Valedictorian

Posted on

Since the terms are in AP, the common difference between the successive terms should be equal.

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

9*3^2x -3^x +1 = 5*3^x +1 -9*3^2x

Collect 3^2x and 3^x together.

18*3^2x - (5+1)3^x -2

18t^2 -6t  = 0, where t = 3^x and 3^2x = (3^x)^2 = t^2.

6t(3t-2) = 0

t = 0 or 3t-2 = 0

t = 0 gives: 3^x = 0 or x = -infinity.

3t -2 = 0 gives 3*3^x -2 = 0 or 3^x = 2/3. Or x = log(2/3)/ log3.

We’ve answered 318,979 questions. We can answer yours, too.

Ask a question