Question

Answer the following:

Prove by method of induction log_a xⁿ = n log_ax, x > 0, n ∈ N

Answer 1

Let P(n) ≡ log_axⁿ = n log_ax, x > 0, n ∈ N

Step 1: 

For n = 1, L.H.S. = log_ax

R.H.S. = log_ax

∴ L.H.S. = R.H.S. for n = 1

∴ P(1) is true.

Step 2:

Let us assume that for some k ∈ N, P(k) is true,

i.e., log_ax^k = klog_ax    ...(1)

Step 3:

To prove that P(k + 1) is true, i.e., to prove that

log_ax^k+1 = (k + 1) log_ax

Now, L.H.S. = log_ax^k+1

= log_a(x^k·x)

= log_ax^k + log_ax   ...[∵ log_mab = log_ma + log_mb]

= klog_ax + log_ax   ...[By (1)]

= (k + 1)log_ax

= R.H.S.

∴ P(k + 1) is true.

Step 4:

From all the above steps and by the principle of mathematical induction P(n) is true for all n ∈ N.

i.e., log_axⁿ = n log_ax, (x > 0) for all n ∈ N.