What is smaller: log^2 11 or log 12?

We'll write (log 11)^2 = (log
10*1.1)^2

We'll use the product
rule:

log a*b = log a + log b

We'll put
a = 10 and b = 1.1

log 10*1.1 = log 10 + log 1.1
(1)

But log 10 = 1 and we'll substitute it in
(1):

log 10*1.1 = 1 + log 1.1

We'll
square raise:

(log 10*1.1)^2 =  (1 + log
1.1)^2

(1 + log 1.1)^2 > 1 + 2*log
1.1

We'll use the power rule of
logarithms:

2*log 1.1 = log 1.1^2

log
(1.1)^2 = log 1.21

We'll add 1:

1 + log
1.21 = log 10 + log 1.21

log 10 + log 1.21 = log
(10*1.21)

log (10*1.21) = log
12.1

Since the base is bigger than 1, the logarithmic function is
increasing, so, for 12.1>12 => log 12.1 > log
12.

(log 11)^2 > log
12

The smaller number is log
12.