Answer:
Given,
f (x) = loge x ⇒ f (y) = loge y
Consider,
f (xy)
F (xy) = loge (xy)
f (xy) = loge (x × y) [since, logb (a×c) = logb a + logb c]
f (xy) = loge x + loge y
∴ f (xy) = f (x) + f (y)
Answer:
Given,
f (x) = loge x ⇒ f (y) = loge y
Consider,
f (xy)
F (xy) = loge (xy)
f (xy) = loge (x × y) [since, logb (a×c) = logb a + logb c]
f (xy) = loge x + loge y
∴ f (xy) = f (x) + f (y)