Question

Prove by method of induction:
$\rm \log_{a}x^{n}= n \ \log_{a} x, \ x\ \textgreater \ 0, \ n \in N$

Answer 1

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Answer 2

we have to prove that, loga xⁿ = n loga x for all n ∈ N.

we know, a natural number is either odd number or even number.

case 1 : for every, n = k

loga x^k = kloga x .......(1)

case 2 : n is an odd natural number.

so, n = 2k + 1

now, loga x^(2k + 1)

= loga {x^(2k). x^1}

[ we know, p^(m + n) = p^m.p^n]

= loga x^(2k) + loga x

[ applying, log(mn) = logm + logn ]

if case 1 is true then, loga x^(2k) = (2k)loga x is also true.

now, loga x^(2k) + loga x = (2k)loga x + loga x

= (2k + 1)loga x

so, loga x^(2k + 1) = (2k +1)loga x .......(2)

case 3 : n is even natural number

then, n = 2k

so, loga x^n = loga x^(2k)

if case 1 is true then, loga x^(2k) = (2k)loga x is also true.

so, loga x^(2k) = (2k)loga x.......(3)

from equations (1), (2) and (3) it is clear that loga xⁿ = n loga x is correct for all natural value of n.