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Inx (or ln x)
By Brad Rodgers (P1930) on Tuesday, April 25, 2000 - 02:46 am:

In a calculus book I am reading, either I have lost several pages or it doesn't explain what Inx means, but either way I am stuck. Anyway, what does "In" mean. I have picked up from the use of it that if Inx=y, then x=ey, but other than this I am confused. Could someone tell me what In is and how to use it in differentiation?

Thanks,

Brad

By Dave Sheridan (Dms22) on Tuesday, April 25, 2000 - 12:21 pm:

It's not In, but rather ln (ie a lowercase l, rather than an uppercase i). The function you're talking about is called the logarithm, which is where the l comes from. The n refers to natural, since it's related to the e function.

So what's it all about then? First, I'll assume you know what ey means, since you mention it. If not, let me know. So if x=ey, we want to be able to write y= something. Thing is, there's no suitable way of doing this using only +-×÷^! etc (the normal operations), so we introduce the function ln. If you like, you can think of it as shorthand: y=ln(x) simply means x=ey.

If you're not confused already, now consider x=10y. In this case, we'd probably write y=log(x), where the log stands for logarithm to base 10. Base here simply means which number we're taking the power of. Now 10y and ey are generally not the same, so log(x) and ln(x) are not the same function. You can use other bases but there are good reasons why these are the only two considered. It is always possible to specify the logarithm to any base in terms of the logarithm to any other base.

To figure out any properties of log (which hold for any base), you need to reverse things. Think of what ex does. Well, ex × ey = e(x+y). That means ln(xy)=ln(x) + ln(y). And so on.

Finally, to differentiate you can again consider things the other way around.
x=ey so 1=ey×dy/dx, which we can rewrite as dy/dx=1/x.
Note that x-1 is the only power of x we couldn't integrate before introducing logarithms, so it's very helpful to be able to do this.

Hope that helps.

-Dave

By Brad Rodgers (P1930) on Wednesday, April 26, 2000 - 01:49 am:

So far, I understand what you've said. However, my book goes farther to say that if y=ax, then lny=xlna. This has me confused first by its missing introduction, and then by the fact that it says the derivitive of y=exlna is lna exlna. It would seem to me that if the derivitive of ex is ex, then that of exlna would be exlna. How does one get either of these results.

Confused,

Brad

By Sean Hartnoll (Sah40) on Wednesday, April 26, 2000 - 10:31 am:


The first property is related to the result (ea)x = exa.

Hence xa = (elnx)a = ealnx, then taking logs on both sides

ln(xa) = a ln(x), which was the first result.

Now in your second question what is really being differentiated here is the function y=ax = eln(ax) = exlna, by the result we just derived.

Now to differentiate this we need to use the chain rule, which I don't know if you know yet, but it says that the derivative of y=f(g(x)) is
dy/dx = df/dg × dg/dx.
So in this case f = eg and g = xlna, the derivative is
dy/dx = e^g lna, so

dy/dx = lna exlna = lna × ax.

Your mistake is interesting though, it shows that ax does not behave as nicely as ex when you differentiate it, which is why ex is the most commonly used. Also note that the new formula can be reduced to the special ex case, because ln(e) = 1.

Hope this helps,

Sean

By Brad Rodgers (P1930) on Thursday, April 27, 2000 - 05:01 am:

I am still not sure how to obtain the result in the first three lines in your post. I get from my own work on this that ax=eeax. I have a feeling I may be wrong though. If you don't see how I got this, I can try to explain it. I think what I don't see is how a=elna. In your post you used xa which I didn't know from where it came- this may be my problem. Anyway, I think I understood the bottom part of your of your post.

Less confused,

Brad

By Sean Hartnoll (Sah40) on Thursday, April 27, 2000 - 12:22 pm:

It is just the definition of ln, that Dave gave above.
We say y = lnx if ey = x, so it follows by definition that elnx = x or indeed that ln(e^x)=x.

Think of lnx as the inverse function to ex, just like sqrtx is the inverse of x2 (so compare the above results to (sqrtx)2=x and sqrt(x2)=x.

By Brad Rodgers (P1930) on Thursday, April 27, 2000 - 09:38 pm:

Thanks, I believe I am understanding ln better. You have shown quite well all of the different steps to be taken.Thanks once again.

Not confused,

Brad