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