Note: Given surds are of order 2, 3, 4 whose 1 c. m. is 12 Changing each one of given surds to that of order 12 we get : `sqrt(2) = 2^(1/2) = 2^(1/2 xx 6/6)` `= (2^6)^(1/12) = (64)^(1/12)` `= 3sqrt(4) = 4^(1/3) = 4^(1/3 xx 4/4)` `= 4^(4/12) = (4^4)^(1/12)` `= (256)^(1/12)` `= 4sqrt(6) = 6^(1/4)` `= 6^(1/4 xx 3/3) = 6^(3/12)` `= (6^3)^(1/12) = (216)^(1/12)` Now,` (64)^(1/12) < (216)^(1/12) < (256)^(1/12)` `:. sqrt(2) < 4sqrt(6) < 3sqrt(4).`

Note: Clearly, m = 11 and n = 2, `:. (m - 1)^(n + 1) = (11 - 1)^3 = 10^3 = 1000.`

Note: `[(3^m^2)/((3^m)^2)]^1/m = 81 => ((3^m^2)/(3^2m))^(1/m) = 3^4` `or (3^(m^2 - 2m)^(1/m)` `= 3^4 or 3^m(m - 2) . 1/m = 3^4` `or 3^(m - 2) = 3^4 or m - 2 = 4 or m = 6`

Note: On diving, we get` (1 - x^8)/(1 - x^4) ` `= 65/64` `or ((1 - x^4) (1 + x^4))/((1 - x^4))` `= 65/64 or 1 + x^4` `= 65/64` `:. x^4 = (65/64 - 1)` `= 1/64 or x = (1/2^6)^(1/4)` `= + 1/(2^(3/2))` - `= + 1/(2sqrt(2))` -

Note: Given equations are : `2^a + 3^b = 17, 2^2 - 3. 3^b = 5` `or x + y = 17 & 4x - 3y = 5,` where `x = 2^a & y = 3^b.` On solving, we get : x = 8 and y = 9. `:. 2^a = 8 = 2^3` and` 3^b = 9 = 3^2` :. a = 3 and b = 2