If log32, log3(2x - 5), log3(2x - 7/2) are in arithmetic progression,

then the value of x is

equal to

Select one:

Ⓒa. 1

O b. 5

c. 7

O d. 3 or 2

Question

Answer 1

Hi
Taking base as 3
Log3 2 +Log3 (2x-7/2) = 2log3 (2x-5)
we get 2(2x-7/2) = (2x-5)^2
4x-7=4x^2+25-20x
4x^2-24x+32=0
x^2-6x+8=0
(x-4)(x-2)=0
but x cannot be 2 as x has to be greater than 7/2
so x=4

Answer 2

Hi
Taking base as 3
Log3 2 +Log3 (2x-7/2) = 2log3 (2x-5)
we get 2(2x-7/2) = (2x-5)^2
4x-7=4x^2+25-20x
4x^2-24x+32=0
x^2-6x+8=0
(x-4)(x-2)=0
but x cannot be 2 as x has to be greater than 7/2
so x=4

Answer 3

I got X=4 or X=2

Answer 4

I got X=4 or X=2

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If log32, log3(2x - 5), log3(2x - 7/2) are in arithmetic progression,then the value of x isequal toSelect one:Ⓒa. 1O b. 5c. 7O d. 3 or 2

If log32, log3(2x - 5), log3(2x - 7/2) are in arithmetic progression,

then the value of x is

equal to

Select one:

Ⓒa. 1

O b. 5

c. 7

O d. 3 or 2