190
Higher Math
Solution : m n
p m and q n
p q
a ⋅a = a
mq nq
⎛ nq1 ⎞ = ⎜a ⎟ ⎜ ⎟ ⎝ ⎠ =a =a
⋅a
np nq
⎛ nq1 ⎞ = ⎜a ⎟ ⎜ ⎟ ⎝ ⎠
mq
⎛ nq1 ⎞ ⎜a ⎟ ⎜ ⎟ ⎝ ⎠
np
[Definition 6]
mq + np
mq + np nq
[law 6] [Definition 6]
mq np + nq nq m p + n q
=a The following facts should be carefully noted : (i ) If a x = 1 where a > 0 and a ≠ 1 so x = 0 (ii ) If a x = 1 where a > 0 and x ≠ 0 so a = 1 (iii ) If a x = a y where a > 0 and a ≠ 1 so x = y a (iv ) If a x = b x where > 0 and x ≠ 0 so a = b b Example 8. Simplify : If a x = b, b y = c and c z = a show that, xyz = 1 . Solution : Given condition, b = a x , c = b y and a = c z Now, b = a x = (c z ) x = c zx = (b y ) zx = b xyz ⇒ b = b xyz ⇒ b1 = b xyz ∴ xyz = 1 . (Proved) a
a
−1 ⎛ a ⎞b Example 9. If a = b show that, ⎜ ⎟ = a b ; and also prove that of a = 2b if ⎝b⎠ b = 2. Solution : Given a b = b a b
a
1
b
∴ b = (a b ) a = a a ⎛ L.H.S. = ⎛⎜ a ⎞⎟ = ⎜ ab ⎜⎜ ⎝b⎠ ⎝ aa a b
a
a
a
⎞b ⎛ b − ⎞b ⎟ ⎜ a1 ⋅ a a ⎟ = ⎟⎟ ⎜ ⎟ ⎝ ⎠ ⎠ a
= a b ⋅ a −1 = a b Again of a = 2b
−1
R.H.S. (proved)