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is clearly true. Next we assume P (k ) is true, that is log xk = k log(x) and try to
show P (k + 1) is true. Using the Product Rule for Logarithms along with the induction
hypothesis, we get log xk+1 = log xk · x = log xk + log(x) = k log(x) + log(x) = (k + 1) log(x)
Hence, log xk+1 = (k + 1) log(x). By induction log (xn ) = n log(x) is true for all x > 0
and all natural numbers n ≥ 1.
3. Let A be an n × n lower triangular matrix. We proceed to prove the det(A) is the product of
the entries along the main diagonal by inducting on n. For n = 1, A = [a] and det(A) = a,
so the result is (trivially) true. Next suppose the result is true for k × k lower triangular
matrices. Let A be a (k + 1) × (k + 1) lower triangular matrix. Expanding det(A) along the
ﬁrst row, we have
n det(A) = a1p C1p
p=1 Since a1p = 0 for 2 ≤ p ≤ k + 1, this simpliﬁes det(A) = a11 C11 . By deﬁnition, we know that
C11 = (−1)1+1 det (A11 ) = det (A11 ) where A11 is k × k matrix obtained by deleting the ﬁrst
row and ﬁrst column of A. Since A is...
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