# Inequality with Logarithms

The problem is to prove the following inequality:

log_{π}3 + log_{3}π > 2.

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Copyright © 1996-2018 Alexander Bogomolny

### Solution

So, the problem is to prove the following inequality:

log_{π}3 + log_{3}π > 2.

There are just two pieces of information that are important to solving this problem. That is, the knowledge of the definition of *logarithms* is being assumed.

However, just in case, log_{a}b is the power to which a is to be raised to give b:

a^{logab} = b.

If log_{a}b = c, i.e., if a^{c} = b, then a = b^{1/c} which tells us that b to the power of 1/c gives a. In other words, log^{b}a = 1/c = 1/log_{a}b.

It follows that log_{a}b and log_{b}a are reciprocal numbers so that their product is exactly 1:

log_{a}b × log_{b}a = 1.

Also, from the definition of the logarithm it follows that log_{a}b = 1 only if

Now, let's return to the problem:

log_{π}3 + log_{3}π > 2.

So, you should know two facts. First, whatever else can be said of the famous constant π, the only important thing that is relevant here is that _{π}3 ≠ 1 ≠ log_{3}π.

x + 1/x ≥ 2,

with the equality only if x = 1. In general, (a + b)/2 ≥ √ab, with equality only when

Of course, it could have been proved directly. Since x is assumed positive, the inequality obtained by multiplying by,dividing by,adding to both sides,multiplying by x, i.e.,

Now, the solution to the problem becomes transparent. Since log_{π}3 and log_{3}π are two reciprocals, their sum is never less than 2 and would only be equal to 2 in case each of them were 1, but, as we saw, they are not. The problem is solved.

**Note**: it is clear from the solution that appearance of π in the problem is a **red herring**. Replacing π with any positive number different from 3 would give the same result and in exactly the same manner.

### What Is Red Herring

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- Inequality with Logarithms
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- Simultaneous Diameters in Concurrent Circles
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- Schur's Inequality
- Further Properties of Peculiar Circles
- Inequality with Csc And Sin
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- An Inequality for the Cevians through Spieker Point via Brocard Angle
- An Inequality In Triangle and Without
- Problem 3 from the EGMO2017
- Mickey Might Be a Red Herring in the Mickey Mouse Theorem
- A Cyclic Inequality from the 6th IMO, 1964
- Three Complex Numbers Satisfy Fermat's Identity For Prime Powers
- Probability of Random Lines Crossing
- Planting Trees in a Row
- Two Colors - Three Points

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Copyright © 1996-2018 Alexander Bogomolny

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