# Math Help - logarithms **URL:** \ **Category:** Factual Questions\ **Created:** [October 17, 2001, 12:52am UTC](https://boards.straightdope.com/t/math-help-logarithms/87862 "2001-10-17T00:52:47Z")\ **Posts on this page:** 20\ **Page:** 1
**Author:** ![Lockz](https://avatars.discourse-cdn.com/v4/letter/l/f4b2a3/32.png) [@Lockz](https://boards.straightdope.com/u/Lockz)\ **Post date:** [October 17, 2001, 12:52am UTC](https://boards.straightdope.com/t/math-help-logarithms/87862/1 "2001-10-17T00:52:47Z")
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!! ---
**Author:** ![Googler](https://avatars.discourse-cdn.com/v4/letter/g/7993a0/32.png) [@Googler](https://boards.straightdope.com/u/Googler)\ **Post date:** [October 17, 2001, 1:00am UTC](https://boards.straightdope.com/t/math-help-logarithms/87862/2 "2001-10-17T01:00:20Z")
I dont think you can expand ln(a-b) IIRC. ---
**Author:** ![keeper0](https://avatars.discourse-cdn.com/v4/letter/k/b19c9b/32.png) [@keeper0](https://boards.straightdope.com/u/keeper0)\ **Post date:** [October 17, 2001, 1:00am UTC](https://boards.straightdope.com/t/math-help-logarithms/87862/3 "2001-10-17T01:00:42Z")
I do know that ln(y^x) = x_ln(y) and ln(x_y) = 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. ---
**Author:** ![hajario](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/hajario/32/171_2.png) [@hajario](https://boards.straightdope.com/u/hajario)\ **Post date:** [October 17, 2001, 1:02am UTC](https://boards.straightdope.com/t/math-help-logarithms/87862/4 "2001-10-17T01:02:17Z")
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 ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [October 17, 2001, 1:04am UTC](https://boards.straightdope.com/t/math-help-logarithms/87862/5 "2001-10-17T01:04:10Z")
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). ---
**Author:** ![Lockz](https://avatars.discourse-cdn.com/v4/letter/l/f4b2a3/32.png) [@Lockz](https://boards.straightdope.com/u/Lockz)\ **Post date:** [October 17, 2001, 1:26am UTC](https://boards.straightdope.com/t/math-help-logarithms/87862/6 "2001-10-17T01:26:17Z")
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. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [October 17, 2001, 1:32am UTC](https://boards.straightdope.com/t/math-help-logarithms/87862/7 "2001-10-17T01:32:54Z")
So what was the original equation? ---
**Author:** ![Lockz](https://avatars.discourse-cdn.com/v4/letter/l/f4b2a3/32.png) [@Lockz](https://boards.straightdope.com/u/Lockz)\ **Post date:** [October 17, 2001, 1:43am UTC](https://boards.straightdope.com/t/math-help-logarithms/87862/8 "2001-10-17T01:43:25Z")
The original equation was (1-(-7)^(k+1))/4 + 2(-7)^(k+1) = (1-(-7)^(k+2))/4 ---
**Author:** ![Trucido](https://avatars.discourse-cdn.com/v4/letter/t/ea5d25/32.png) [@Trucido](https://boards.straightdope.com/u/Trucido)\ **Post date:** [October 17, 2001, 1:48am UTC](https://boards.straightdope.com/t/math-help-logarithms/87862/9 "2001-10-17T01:48:10Z")
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. ---
**Author:** ![barking\_frog](https://avatars.discourse-cdn.com/v4/letter/b/96bed5/32.png) [@barking\_frog](https://boards.straightdope.com/u/barking_frog)\ **Post date:** [October 17, 2001, 1:49am UTC](https://boards.straightdope.com/t/math-help-logarithms/87862/10 "2001-10-17T01:49:08Z")
No k satisfies that equation. ---
**Author:** ![Trucido](https://avatars.discourse-cdn.com/v4/letter/t/ea5d25/32.png) [@Trucido](https://boards.straightdope.com/u/Trucido)\ **Post date:** [October 17, 2001, 1:50am UTC](https://boards.straightdope.com/t/math-help-logarithms/87862/11 "2001-10-17T01:50:44Z")
Oh crap. You can’t do that. I was wondering why it was so easy. Well, I’m an idiot. ---
**Author:** ![erislover](https://avatars.discourse-cdn.com/v4/letter/e/71e660/32.png) [@erislover](https://boards.straightdope.com/u/erislover)\ **Post date:** [October 17, 2001, 1:57am UTC](https://boards.straightdope.com/t/math-help-logarithms/87862/12 "2001-10-17T01:57:36Z")
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. ---
**Author:** ![Lockz](https://avatars.discourse-cdn.com/v4/letter/l/f4b2a3/32.png) [@Lockz](https://boards.straightdope.com/u/Lockz)\ **Post date:** [October 17, 2001, 1:59am UTC](https://boards.straightdope.com/t/math-help-logarithms/87862/13 "2001-10-17T01:59:21Z")
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 ---
**Author:** ![Lockz](https://avatars.discourse-cdn.com/v4/letter/l/f4b2a3/32.png) [@Lockz](https://boards.straightdope.com/u/Lockz)\ **Post date:** [October 17, 2001, 2:01am UTC](https://boards.straightdope.com/t/math-help-logarithms/87862/14 "2001-10-17T02:01:34Z")
And, of course, **erislover** comes along and shows the answer TWO MINUTES before my reply appears. ☹ Oh well, no biggie. Thanks! ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [October 17, 2001, 2:02am UTC](https://boards.straightdope.com/t/math-help-logarithms/87862/15 "2001-10-17T02:02:39Z")
> [@](#): > > \*Originally posted by Lockz \* > \*\*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 \*\* Just bear in mind that that’s not a proof, though, as it starts from the assumption that both sides are equal. ---
**Author:** ![barking\_frog](https://avatars.discourse-cdn.com/v4/letter/b/96bed5/32.png) [@barking\_frog](https://boards.straightdope.com/u/barking_frog)\ **Post date:** [October 17, 2001, 2:05am UTC](https://boards.straightdope.com/t/math-help-logarithms/87862/16 "2001-10-17T02:05:37Z")
Oh crap, didn’t see the two as the exponent on the RS… typed too soon I guess. ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [October 17, 2001, 2:15am UTC](https://boards.straightdope.com/t/math-help-logarithms/87862/17 "2001-10-17T02:15:02Z")
> [@](#): > > \*Originally posted by ultrafilter \* > \*\*Just bear in mind that that’s not a proof, though, as it starts from the assumption that both sides are equal. \*\* Of course, **erislover** did mention that too… ---
**Author:** ![g8rguy](https://avatars.discourse-cdn.com/v4/letter/g/7cd45c/32.png) [@g8rguy](https://boards.straightdope.com/u/g8rguy)\ **Post date:** [October 17, 2001, 3:12am UTC](https://boards.straightdope.com/t/math-help-logarithms/87862/18 "2001-10-17T03:12:28Z")
> [@](#): > > \*Originally posted by Googler \* > \*\*I dont think you can expand ln(a-b) IIRC. \*\* 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 ---
**Author:** ![ultrafilter](https://avatars.discourse-cdn.com/v4/letter/u/3d9bf3/32.png) [@ultrafilter](https://boards.straightdope.com/u/ultrafilter)\ **Post date:** [October 17, 2001, 3:45am UTC](https://boards.straightdope.com/t/math-help-logarithms/87862/19 "2001-10-17T03:45:40Z")
> [@](#): > > \*Originally posted by g8rguy \* > \*\* > > > [@](#): > > > > \*Originally posted by Googler \* > > \*\*I dont think you can expand ln(a-b) IIRC. \*\* > > 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? ---
**Author:** ![g8rguy](https://avatars.discourse-cdn.com/v4/letter/g/7cd45c/32.png) [@g8rguy](https://boards.straightdope.com/u/g8rguy)\ **Post date:** [October 17, 2001, 5:26am UTC](https://boards.straightdope.com/t/math-help-logarithms/87862/20 "2001-10-17T05:26:15Z")
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… [Next page](https://boards.straightdope.com/t/math-help-logarithms/87862.md?page=2)