If a>b, lna>lnb
The statement is true: see OP, x>0. And anyhow, lna only defined for a positive. It is a strictly increasing function because either by def of lnx as integral of 1/t from 1 to x, x>0, or most easily by d(logx)/dx = 1/x and x>0. see OP.
ie, lna>lnb if a>b because derivative is pos.
Why make everything so complicated?
I give a 3 line answer to the OP and you turn a mole-hill into a mountain.
I did not make everything so complicated. I answered the OP's second question (which you ignored to start a rant that proofs don't need supporting arguments). You can read the OP's second question here. It is rude to hijack threads and troll those attempting to actually help.
I proved OP with
x>in(1+x)
e^x>1+x
e^x = 1+x+.... >1+x
Every step is trivial and would have been taken for granted in an introductory calculus course.
But I was asked to show: e^x>e^y -> x > y which I did by showing that
If a>b, lna>lnb which I considered trivial at this level of discussion (obviously derivative of ln>0 without a lot of blah-blah, or look at the graph). Actually, this step was hi-jacked by someone else with an exagerated discussion in abstract terms to prove this trivial statement.
What did I hi-jack? One of the other posts? you’re kidding.
And yet a statement like substitute into Napier's inequity and the answer falls right out goes unquestioned.
I was not going to respond to you the first time that you posted this. Then, as I continued replying to the OP, you made the exact same post a second time and asked, "what's wrong with this?". I mistook it for a genuine question. Obviously my mistake. I apologize for whatever I said that put you on the defensive.
Thank you. This was constructive to the thread.
Napier invented logarithms. Not only is Napier's inequality pertinent, it is also very interesting that the answer does fall right out. I provided one method of proof, Plato provided a second, and you provided a third (although I still think the order of your three statements should be reversed). Again, the problem started because I mistook what you said as a question. It is difficult to read inflection in a post. I apologize once more for putting you on the defensive.
You are correct. Three sols were given. The first in the first post.
1) f(x) = x-ln(1+)x and f’(x)>0. Good answer.
2) Napier’s inequality. Answer
My only objection to this answer is it proves one unproved inequality with another unproved inequality. Couldn’t find any reference to it in a calculus book. I vaquely recall seeing it in statistics, so possibly inappropriate to subject. So the answer depends on happening to remember a formula from somewhere.
3) x>lnx, e^x>1+x, e^x=1+x+…>1+x. Answer
I missed 1) and 2) because I was thrown off by the endless pages of math which I had no inclination to follow and which led me to believe that the respondents were still fishing for an answer.
3) was posted after a considerable lull and then followed soon after by another huge chunk of math out of the clear blue sky in response to an unreferenced “hint.” I could only conclude without wasting a lot of time that the answer was still being sought, which puzzled me since an answer had just been given.
The suggestion that I hi-jacked an answer is outrageous. All three answers are clearly independent.
Whoops. 1) is not the answer. f'(x)>0 means f(x) increasing but not necessarily positive. So you have to show it's positive, ie, x>ln(1+x), for x -> 0. Or did I miss something? Was that the point of all those pages of math? Or was there another answer buried somewhere else?
That really only leaves 3) in the spirit of the question which can be shortened to:
e^x=1+x+…>1+x
EDIT: OK, and the inverse (ln) of an increasing function is increasing. I like the essence of a solution without distractions. That's why I liked 1) before I realized it wasn't the essence of the solution. But at least it was clear.