Logarithms Proof

[govindpreetsingh]

if a > 0, b > 0, and c > a+b

prove that: log a + log b <= 2log c - 2

[Math Tutor]

What is the logarithm base?

[govindpreetsingh]

Hello

Well the log base could be any.

Actually, I was able to prove it using the following:

a+b <= c

dividing both the sides by 2, we get

(a+b)/2 <= c/2

logging both the sides, we get

log (a+b)/2 <= log (c/2)

1/2 (log a + log b) <= log c - log 2

1/2 (log a + log b) <= log c - 1 [because with reference to log table, log 2 = 1]

log a + log b <= 2(log c - 1)

log a + log b <= 2logc - 2

Hence Proved!

Please correct me if I am wrong anywhere.
Thanks a ton and I will keep posting more questions here as they would arise.

Govind Preet Singh

[Math Tutor]

The proof is OK, except this:

(a+b)/2 <= c/2

logging both the sides, we get

log (a+b)/2 <= log (c/2)

It is true when the logarithm base > 1
If it is smaller than 1
the equation sign is ≥ - the opposite

I have to confess that the problem is very interesting!

[govindpreetsingh]

Hello Again,

1) this proof is assumed to be using a base 2
2) on a very first look, it does look resolved, but there is a fundamental error in this proof, and that would be:

log (a+b/2) ≠ 1/2 (log a + log b)
so my proof is wrong above - yes I admit it myself.
[Unfortunately I discovered it this morning]

please note that there is a log rule that says:

if a>0, b>0
then (a+b/2)² ≤ √ab

so using the above statement, we need to prove this theorem.

If we simplify the theorem statement, it comes to:
log₂ a + log₂ b ≤ log₂ c² - 2

So this is what we will have to prove.

[Math Tutor]

My proof
c² > (a + b)² ≥ 4ab

[tex]log_2c^2 > log_24ab[/tex]
[tex]2log_2c > log_24 + log_2a + log_2b[/tex]

[tex]2log_2c - 2 > log_2a + log_2b[/tex]


If you do not know why (a + b)² ≥ 4ab

a² + 2ab + b² ≥ 4ab

a² - 2ab + b² ≥ 0

(a - b)² ≥ 0