Question

Let f(x) = log(x). Find values of a such that f(kaa) = kf(a).

Answer 1

Answer:

a = $k^{\frac{1}{k-2}}$

Step-by-step explanation:

Given:

f(x) = log(x)

and,

f(kaa) = kf(a)

now applying the given function, we get

⇒ log(kaa) = k × log(a)

or

⇒ log(ka²) = k × log(a)

Now, we know the property of the log function that

log(AB) = log(A) + log(B)

and,

log(Aᵇ) = b × log(A)

Thus,

⇒ log(k) + log(a²) = k × log(a)         (using log(AB) = log(A) + log(B) )

or

⇒ log(k) + 2log(a) = k × log(a)            (using log(Aᵇ) = b × log(A) )

or

⇒ k × log(a) - 2log(a) = log(k)

or

⇒ log(a) × (k - 2) = log(k)

or

⇒ log(a) = (k - 2)⁻¹ × log(k)

or

⇒ log(a) = $\log(k^{\frac{1}{k-2}})$          (using log(Aᵇ) = b × log(A) )

taking anti-log both sides

⇒ a = $k^{\frac{1}{k-2}}$