Question

Show that (a) log(1 + 2 + 3) = log 1 + log 2 + log 3 (b) log(1 + ac) = 2 log b, where a,
b and c are consecutive numbers.

Answer 1

Question :-

1. Show that

(a) log(1 + 2 + 3) = log 1 + log 2 + log 3

(b) log(1 + ac) = 2 log b, where a, b and c are consecutive numbers.

Answer

Formula used :-

$\bf \: log(xy) = logx + logy$

$\bf \: log(\dfrac{x}{y} ) = logx - logy$

$\bf \:log {x}^{y} = ylogx$

Solution :-

Part (a) :-

$\bf \:Consider \: log(1 + 2 + 3)$

$\bf\implies \: log(6)$

$\bf\implies \: log(1 \times 2 \times 3)$

$\bf\implies \: log(1) + log(2) + log(3)$

Part (b) :-

Let three consecutive numbers be x, x + 1, x + 2

So, a = x, b = x + 1, c = x + 2.

$\bf \:Consider \: log(1 + ac)$

$\bf\implies \: log(1 + x(x + 2))$

$\bf\implies \: log(1 + {x}^{2} + 2x )$

$\bf\implies \: log {(x + 1)}^{2}$

$\bf\implies \:2 log(x + 1)$

$\bf\implies \:2 log(b)$

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