`(a/b)^(x-1)` =` (b/a)^(x-3)`   `hArr`  `(a/b)^(x-1)` =  `(a/b)^-(x -3)`   = `(a/b)^(3-x)`

                                     `hArr`  `x - 1 = 3 - x    hArr   2x = 4   hArr     x= 2`

`((243)^0.13 xx (243)^0.07)/((7)^0.25 xx (49)^0.075 xx (343)^0.2)`

= `((243)^(0.13 + 0.07))/(7^0.25 xx (7^2)^0.075 xx (7^3)^0.2)`

= `((243)^0.2)/(7^0.25 xx 7^(2 xx 0.075) xx 7^(3 xx 0.2))`  = `((3^5)^0.2)/(7^0.25 xx 7^0.15 xx 7^0.6)`

 =`(3^((5 xx 0.2)))/(7^((0.25 + 0.15 + 0.6)))` = `3^1/7^1` =  `3/7`

Let `(25)^7.5 xx (5)^2.5 -: (125)^1.5` = `5^x`. Then ,`((5^2)^7.5 xx (5)^2.5)/((5^3)^1.5)` =  `5^x`

`hArr` ` (5^(2 xx 7.5) xx 5^2.5)/( 5^(3 xx 1.5))`  = `5^x`

`hArr` ` (5^15 xx 5^2.5)/(5^4.5)` = `5^x`  `hArr`  `5^x` = `5^(15 + 2.5 - 4.5)`  = `5^13 `      `hArr`   `x` =  13.

`(18)^3.5 -: (27)^3.5  xx 6^3.5`   =  `2^x`

`hArr`    `(18)^3.5  xx (1)/(27)^3.5 xx 6.35` = `2^x`  ` hArr`   `(3^2 xx 2)^3.5 xx (1)/(3^3)^3.5  xx (2 xx 3)^3.5`  = `2^x`

`hArr`     `3^((2 xx 3.5))  xx 2^3.5 xx  1/3^((3 xx 3.5)) xx 2^3.5 xx 3^3.5`  =`2^x`

`hArr`     `3^7 xx 2^3.5 xx 1/3^10.5 xx 2^3.5 xx 3^3.5` = `2^x`      `hArr` ` 2^7` = `2^x`   `hArr`     `x` =  7.

`(64)^(-(1)/(2)) - (- 32)^(-(4)/(5)) `= `(8^2)^(-(1)/(2)) - [(-2)^5]^(-(4)/(5))`      = `8^(2 xx ((-1))/2)`  -` (- 2)^(5 xx ((- 4))/5)`

= `8^-1 - ( - 2)^-4`=  `1/8 -  1/(-2)^4` =  `(1/8 - 1/16)` =  `1/16`

`8^-25 - 8^-26` = `((1)/(8^25) - (1)/(8^26))` = `((8 - 1))/(8^26) ` = ` 7 xx 8^-26`.

`49 xx 49 xx 49 xx 49` = `(7^2 xx 7^2 xx 7^2 xx 7^2)` = `7^((2 + 2 + 2 + 2))` = `7^8`

So the correct answer is 8.

Let `(17)^3.5 xx (17)^x` =   `17^8`.  Then `,(17)^(3.5 + x)` = `17^8`

`:.`   `3.5 + x =  8      hArr     x = (8 - 3.5)     hArr     x    = 4.5`

`(0.04)^ 1.5`  = `(4/100)^ - 1.5` = `(1/25)^(-3/2)` = `(25)^(3/2)` = `(5^2)^(3/2)` = `5^((2 xx 3/2))` = `5^3` = 125.

`(256)^0.16 xx (256)^0.09` =  `(256)^((0.16 + 0.09))` =  `(256)^0.25` =  `(256)^(((25)/(100)))`
= `(256)^((1)/(4))` = `(4^4)^((1)/(4))` = `4^((4 xx 1/4))` = `4^1` = 4.