Is Logarithm Inequality True for 0 < a < b < 1?

[l-1j-cho]

if 0 < a < b < 1, is log(a) < log (b) true for all a and b?
I haven't found any counter example but just to make it sure.

 

[I like Serena]

Hi l-1j-cho!

Yes, it is actually true for any 0 < a < b, since the log function is a strictly increasing function.

 

[l-1j-cho]

thank you!

 

[pessimist]

take log b - log a and differentiate it.

 

[Nanas]

It is true for a<b and we know that the domain is (0,infinity) where the base is greater than 1 Because The exponetial function are increasing function with base greater than one and we know that logarithms are inverse functions of exponetial functions try to prove it.

Thanks

 

[SammyS]

It depends upon what you mean by the log function -- what base is being used.

If 1 < G,

then log _G (x) is strictly increasing on its entire domain, (0, +∞).

However, 0 < 1/G < 1 and we have log _1/G (x) is strictly decreasing on its entire domain, (0, +∞). You conjuncture would be false in this case​

So if by the log function, you mean log_e a.k.a. ln, or if you mean log₁₀, then the log function is strictly increasing.

If on the other hand, you are referring to the logarithm function generically, then your conjecture is true if and only if the base is greater than 1.

It is somewhat unusual to use a logarithm with a base in the range (0, 1).

 

[Nanas]

It is somewhat unusual to use a logarithm with a base in the range (0, 1).

I didn't understand what you mean here.Why it is unusual ?

 

[SammyS]

It's just unusual to see something like log _0.2 (6), for example.

If 1 < G, then 0 < 1/G < 1 .

Doing a change of base gives:
$$\log_{\,1/G}\,(x) =\frac{\log\,_G\,(x)}{\log_{\,\,G}\,(1/G)}=-\log_{\,G}\,(x)$$