I’m going through a mathematical induction question (yes, in my homework), when I came across this:
ln(1-(-7)^(k+1))
I learned logarithms a few years ago, but I’ve forgotten how to expand this. Can a Doper help me out please?
Thanks!!
I dont think you can expand ln(a-b) IIRC.
I do know that ln(y^x) = xln(y)
and ln(xy) = ln(x) + ln(y)
but I haven’t the foggiest idea of what do with ln(x+y), which is what you have. My calc book doesn’t list any such expansion.
I think that this can’t be expanded any more than it is. Are you sure all of your parentheses are in the correct place.
Haj
Yeah, ln(x + y) doesn’t have any particular expansion that I’m aware of (and some of the books I have would contain it). However, note that if a is positive, ln(a + b) >= ln(b).
ln(a+b) wasn’t in my calculus text as well, so I thought I’d bring it here to ask people if they had seen something similar. I think I made an error solving for the variable in my equation. I will try to use some other way of solving other than taking the natural log of both sides.
So what was the original equation?
The original equation was
(1-(-7)^(k+1))/4 + 2(-7)^(k+1) = (1-(-7)^(k+2))/4
ln(1-(-7)^(k+1))=(k+1)*ln(1-(-7))=(k+1)ln(8)
It’s just a property of logarithms that you can shift that exponent out.
No k satisfies that equation.
Oh crap. You can’t do that. I was wondering why it was so easy. Well, I’m an idiot.
If you are doing induction work, your best bet is just to simplify the side of the equation which would be of the form:
x + (x+1) = whatever
This can be seen as follows:
you have an equation of the following form:
(1-a[sup]k+1[/sup])/4+4a[sup]k+1[/sup]=(1-a[sup]k+2[/sup])/4
Get your left side to be a single fraction as follows:
(1-a[sup]k+1[/sup])/4+(8a[sup]k+1[/sup])/4 = …
Combine terms
(1+7(a[sup]k+1[/sup]))/4 = …
Now, note that what I called a is -7, so we say instead:
(1-(-7)(-7[sup]k+1[/sup]))/4 = …
and note that a*a[sup]x[/sup] = a[sup]x+1[/sup] and say
(1-(-7)[sup]k+1+1[/sup])/4 = …
And there ya go.
Though it is safe, in testing identities, to perform operations on both side of the equals sign, it isn’t very safe to move stuff around across the equals sign because then that assumes what you are trying to prove. Sort of a thin line— and so your best bet is to try and perform simepl algebraic operation on terms without doing anything involving both sides of the equation.
I just solved it. I didn’t need a k to satisfy the equation, I just needed to show the two sides equal each other. Here’s my proof:
(1-(-7)^(k+1))/4 + 2(-7)^(k+1) = (1-(-7)^(k+2))/4
I’m going to work on the left side of the equation
= (1-(-7)^(k+1)/4 + 4(2(-7)^(k+1))/4
= (1-(-7)^(k+1)/4 + 8(-7)^(k+1))/4
= (1-(-7)^(k+1) + 8(-7)^(k+1))/4
= (1 + 7(-7)^(k+1))/4
= (1 - (-7)(-7)^(k+1))/4
= (1 - (-7)^(k+2))/4
Therefore, LS = RS
Yay, I did it!
Mods, you can please close this, as the question has been solved. Thanks for all your help!
Lockz
And, of course, erislover comes along and shows the answer TWO MINUTES before my reply appears. 
Oh well, no biggie. Thanks!
Just bear in mind that that’s not a proof, though, as it starts from the assumption that both sides are equal.
Oh crap, didn’t see the two as the exponent on the RS… typed too soon I guess.
Of course, erislover did mention that too…
Just to add my scant wisdom to this… Sure you can! You just don’t get anything very helpful. Let’s assume a is bigger than b (otherwise, just switch a and b in what follows, although you’ll have the logarithm of a negative number).
ln(a-b) = ln(a*(1-b/a)) = ln(a) + ln(1-b/a)
ln(1-x) = -[sym]S/sym, where the sum runs on n from 1 to infinity (which is why it’s not so helpful except for numerical approximations).
insert the second line into the first, with x = b/a, and the not very illuminating result is
ln(a-b) = ln(a) - [sym]S/sym
So what’s the interval of convergence for that series?
As far as I can remember, it converges for |x| < 1. I’m no expert on convergence properties of series, though. One of the dirty little secrets of physics is that we tend to leave such trivial little details like that to the mathematicians. 
The series is just the Taylor series about x=0, and the first place I can think of that ln(1-x) blows up is at x=1, for what it’s worth…