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

Question

Answer the following:

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

Sum

Solution

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.

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Principle of Mathematical Induction

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Q II. (11) (i)Q II. (10) (ii)Q II. (11) (ii)

Chapter 4: Methods of Induction and Binomial Theorem - Miscellaneous Exercise 4.2 [Page 86]

APPEARS IN

Balbharati Mathematics and Statistics 2 (Arts and Science) [English] Standard 11 Maharashtra State Board

Chapter 4 Methods of Induction and Binomial Theorem
Miscellaneous Exercise 4.2 | Q II. (11) (i) | Page 86

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