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

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### 2 Answers

Here we have to compare (log 11) ^2 and log 12. We assume the logarithms have the base 10.

Now (log 11) ^2 can be written as (log 10*1.1) ^2

=> (log 10 + log 1.1) ^2

=> (1+ log 1.1) ^2

Now (1+ log 1.1) ^2 = 1+ (log 1.1) ^2 + 2 log 1.1

= (log 1.1) ^2 + 1 + log (1.1) ^2

= (log 1.1) ^2 + log 10 + log 1.21

= (log 1.1) ^2 + log 10*1.21

= (log 1.1) ^2 + log 12.1

Now (log 1.1) ^2 is positive.

Also log 12.1 > log 12

Therefore we can say that (log 11) ^2 > log 12

**The smaller number is 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.**