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

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प्रश्न

Answer the following:

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

योग
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उत्तर

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.

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  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 4: Methods of Induction and Binomial Theorem - Miscellaneous Exercise 4.2 [पृष्ठ ८६]

APPEARS IN

बालभारती Mathematics and Statistics (Arts and Science) Part 2 [English] Standard 11 Maharashtra State Board
अध्याय 4 Methods of Induction and Binomial Theorem
Miscellaneous Exercise 4.2 | Q II. (11) (i) | पृष्ठ ८६

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