Arithmetically Progressive terms

First, we let these variables for simplicity: $a = \log_3 2$ $b = \log_3 (2^x - 5)$ $c = \log_3 (2^x - 3.5)$

Since these three numbers are in Arithmetic Progression, then $b = a + d$ $c = b + d$

where $d$ is the common difference.

If we subtract $b$ from $c$ , then $c - b = b - a$ $2b = a + c$

Substituting back the variables, $2 \log_3 (2^x - 5) = \log_3 2 + \log_3 (2^x - 3.5)$

Using the properties of logarithm, $\log_3 (2^x - 5)^2 = \log_3 2(2^x - 3.5)$ $(2^x - 5)^2 = 2(2^x - 3.5)$

We now let $2^x = e$ for simplicity $(e - 5)^2 = 2(e - 3.5)$ $e^2 - 10e + 25 = 2e - 7$ $e^2 - 12e + 32 = 0$ $(e - 4)(e - 8) = 0$

The roots are $4$ and $8$ . Substituting back $e$ , we will get $2^x = 4$ and $2^x = 8$

We will get $x$ to be $2$ and $3$ . But $2$ is not in the domain of $x$ since if we substitute $2$ for $x$ in $b$ , we will get $\log_3 (- 1)$ . We cannot have a negative value for the argument of the logarithmic function. Thus, the answer is $3$ .

since these 3 are in AP therefore (2 ), ( 2^x -5 ) ,(2^x - 3.5 ) are in GP there fore by applyinf b^2=ac

we get 2^2x - 12*2^x +32=0

therefore x=2 , 3 but x is not equal to 2 as log(-1)/log 3 is not defined

therefore x=3

Log (2^x/ 2) = Log [(2^x - 3.5)/ (2^x - 5)]

(2^x)^2 - 12 (2^x) + 32 = 0

(2^x - 8)(2^x - 4) = 0

2^x = 8 or 2^x = 4

x = 3 or x = 2

Since 2^x - 5 < 0 with x = 2 for negative domain not preferred,

x = 3 {Answer preferred is usually answer wanted.}

Note:

For L.H.S. = R.H.S

=> Log -1 - Log 2 = - Log 2 - Log -1 is perfectly correct as well.

As - Log -1 = Log (1/ -1) = Log -1 also

LN 2 = Ln 2 + j a Pi and LN -1 = j b Pi

Where a = {... , -4, -2, 0, 2, 4, ...} and b = {... , -3, -1, 1, 3, ...}

It is just that A.P. does not take complex numbers which do not follow an order of greater or smaller.

it is the basic property that if log(a) ,log(b) ,log (c) are in a.p. ,,,,then a,b,c are in g.p. Now for g.p. we have b^2 =ac SO (2^x-5)^2=2(2^x-3.5) by solving quadratically we get 2^x=8,4 so values came out to be 3 and 2 but 2 is not in the domain of it because 2^2-5=-1 .AND log(-1 )is not possible So ANSWER IS "3".