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=e^y , but other than this I am confused. Could someone tell me what In is and how to use it in differentiation?

Thanks,

Brad

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 e^y means, since you mention it. If not, let me know. So if x=e^y , we want to be able to write y= something. Thing is, there's no suitable way of doing this using only + - x / ^ ! 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=e^y .

If you're not confused already, now consider x=10^y . 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 10^y and e^y 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 e^x does. Well, e^x xe^y = 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=e^y so 1=e^y x 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

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

Confused,

Brad

Sean

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 a^x =e^eax . 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=e^lna . In your post you used x^a 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

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

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

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