# Answer the following: Prove by method of induction loga xn = n logax, x > 0, n ∈ N - Mathematics and Statistics

Sum

Prove by method of induction loga xn = n logax, x > 0, n ∈ N

#### Solution

Let P(n) ≡ logaxn = n logax, x > 0, n ∈ N

Step 1:

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

R.H.S. = logax

∴ 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., logaxk = klogax    ...(1)

Step 3:

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

logaxk+1 = (k + 1) logax

Now, L.H.S. = logaxk+1

= loga(xk·x)

= logaxk + logax   ...[∵ logmab = logma + logmb]

= klogax + logax   ...[By (1)]

= (k + 1)logax

= 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., logaxn = n logax, (x > 0) for all n ∈ N.

Concept: Principle of Mathematical Induction
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#### APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) 11th Standard Maharashtra State Board
Chapter 4 Methods of Induction and Binomial Theorem
Miscellaneous Exercise 4 | Q II. (11) (i) | Page 86