Note: Given Exp.`q/(q + pq + 1) + r/(r + qr + 1) + p/(p + pr 1)` `= q/(q + pq + 1) + 1/(1 + q + pq) + pq/(pq + 1 q) [:. r = 1/pq]` `= (q + 1 + pq)/(q + 1 + pq)` `= 1`

Note: Given Exp.`x^((a - b)/(ab)) . x^((b - c)/(ab)) . x^((c - a)/(ca))` `= x^(a^2 - b^2) + (b^2 - c^2) + (c^2 - a^2)` `= x^0 = 1.`

Note: Not available

Note: `(a/b)^(x - 1) = (b/a)^(x - 3) <=> (a/b)^(x - 1) = (a/b)^(3 - x)` `:. x - 1 = 3 - x or 2x = 4 or x = 2`

Note: `2^x xx 8^(1/5) <=> (2^x xx 2^(3/5))/(2^(1/5)` `= 1 <=> 2^x xx 2(3/5 - 1/5) = 1` `or 2(x + 2/5) = 2^0.` So,` x + 2/5 = 0 or x = - 2/5`

Note: On squaring both sides, we get : `5 + sqrt(x) = 9 or 3sqrt(x) = 4` Cubing both sides, we get `x = (4 xx 4 xx 4) = 64`

Note: `5^(x + 3) = (25)^(3x - 4 ) `or 5^(x + 5) = 5^(2(3x - 4) or 5^(x + 5)` `= 5^(6x - 8)` `:. x + 3 = 6x - 8 or 5x = 11 or x = 11/5.`

Note: `3sqrt(32) = 2^x => (2^5)^(1/5)` `= 2^x => 2^x = 2^(5/3)` `:. x = 5/3`

Note: Let` a^2 = b^y = c^z = k` Then `a = k^(1/x), b = k^(1/y)` and `c = k^(1/z)` `b^2 = ac => k^(2/y)` ` = k^(1/z) => k^(2/y) = k^(1/x + 1/z)` `:. 2/y = 1/x + 1/z or 2/y` `= (x + z)/(xz) or y = (2xz)/(x + z)`

Note: `2^x = 3^y = 6^-z` `= k => 2 = k^(1/x), 3 = k^(1/y) & 6 = k^(- 1/z)` Now `2 xx 3 = 6 => k^(1/x) xx k^(1/y)` `= k^(- 1/2) => k(1/x + 1/y)` `= k^(- 1/z)` `:. 1/x + 1/y = - 1/z or 1/x + 1/y + 1/z = 1`