# Dumb-ass algebra question **URL:** \ **Category:** Factual Questions\ **Created:** [February 22, 2004, 12:38am UTC](https://boards.straightdope.com/t/dumb-ass-algebra-question/230437 "2004-02-22T00:38:05Z")\ **Posts on this page:** 12\ **Page:** 1
**Author:** ![bump](https://avatars.discourse-cdn.com/v4/letter/b/7c8e57/32.png) [@bump](https://boards.straightdope.com/u/bump)\ **Post date:** [February 22, 2004, 12:38am UTC](https://boards.straightdope.com/t/dumb-ass-algebra-question/230437/1 "2004-02-22T00:38:05Z")
Ok… I’m having an algebretarded moment… If I have 1.1^x / 1.2^x = 5/7, how do I solve for x? (and no, I’m not a high school student doing his homework! I’m a grad student working on his homework who can’t remember his algebra from 13 years ago) ---
**Author:** ![tremorviolet](https://avatars.discourse-cdn.com/v4/letter/t/e47774/32.png) [@tremorviolet](https://boards.straightdope.com/u/tremorviolet)\ **Post date:** [February 22, 2004, 12:43am UTC](https://boards.straightdope.com/t/dumb-ass-algebra-question/230437/2 "2004-02-22T00:43:16Z")
OK, I’m gonna feel _really_ stupid if I’m wrong but I dont’ think that’s algebra. I think yer gona hafta use differential equations to solve that. (and I dont’ remember any of that without a textbook handy but I’m sure a more mathematically gifted Doper wil be along shortly to either tell me I’m wrong or enlighten us…) ---
**Author:** ![tremorviolet](https://avatars.discourse-cdn.com/v4/letter/t/e47774/32.png) [@tremorviolet](https://boards.straightdope.com/u/tremorviolet)\ **Post date:** [February 22, 2004, 12:48am UTC](https://boards.straightdope.com/t/dumb-ass-algebra-question/230437/3 "2004-02-22T00:48:33Z")
Could I _be_ a bigger moron? (completely ignore that first post-ARGH! I want edit!) After thinking about it for half a second: 1.1^x/1.2^x = 5/7 is the same as 1.1^x = 5 and 1.2^x = 7. So break up the equations and solve for x. Should be the exact same number for both equations. ---
**Author:** ![g8rguy](https://avatars.discourse-cdn.com/v4/letter/g/7cd45c/32.png) [@g8rguy](https://boards.straightdope.com/u/g8rguy)\ **Post date:** [February 22, 2004, 12:54am UTC](https://boards.straightdope.com/t/dumb-ass-algebra-question/230437/4 "2004-02-22T00:54:33Z")
> [@tremorviolet](#): > > Could I _be_ a bigger moron? (completely ignore that first post-ARGH! I want edit!) After thinking about it for half a second: > > 1.1^x/1.2^x = 5/7 is the same as 1.1^x = 5 and 1.2^x = 7. So break up the equations and solve for x. Should be the exact same number for both equations. Or you could write 1.1[sup]x[/sup] / 1.2[sup]x[/sup] = (1.1/1.2)[sup]x[/sup] and do exciting things with logarithms to solve for x. ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [February 22, 2004, 12:55am UTC](https://boards.straightdope.com/t/dumb-ass-algebra-question/230437/5 "2004-02-22T00:55:42Z")
You can’t break it into two equation like that. 1.1^x/1.2^x = (1.1/1.2)^x = (11/12)^x Now, as **g8rguy** says, to solve (11/12)^x = 5/7, you need to use logarithms. If you don’t remember how to work with these, you’ll have to remind yourself. log(11/12)^x = log(5/7) x log(11/12) = log(5/7) x = log(5/7) / log(11/12) = 3.8670 If you’re more comfortable saying, instead, x = log[sub]11/12/sub, that’s the same thing. I prefer the first way, though. ---
**Author:** ![Q.E.D](https://avatars.discourse-cdn.com/v4/letter/q/51bf81/32.png) [@Q.E.D](https://boards.straightdope.com/u/Q.E.D)\ **Post date:** [February 22, 2004, 12:59am UTC](https://boards.straightdope.com/t/dumb-ass-algebra-question/230437/6 "2004-02-22T00:59:43Z")
I plugged the equation into MathCad, out of curiosity, and got the same result as **Achernar**. It came out to 3.8669912862235766523 (is it actually rational, or did MathCad round it off?) ---
**Author:** ![gregongie](https://avatars.discourse-cdn.com/v4/letter/g/919ad9/32.png) [@gregongie](https://boards.straightdope.com/u/gregongie)\ **Post date:** [February 22, 2004, 1:02am UTC](https://boards.straightdope.com/t/dumb-ass-algebra-question/230437/7 "2004-02-22T01:02:02Z")
(1.1)^x = 5 (1.2)^x … 7 (1.1/1.2)^x = (5/7) ln (1.1/1.2)^x = ln (5/7) x ln (1.1/1.2) = ln (5/7) x = ln (1.1/1.2)/ln(5/7) ---
**Author:** ![tremorviolet](https://avatars.discourse-cdn.com/v4/letter/t/e47774/32.png) [@tremorviolet](https://boards.straightdope.com/u/tremorviolet)\ **Post date:** [February 22, 2004, 1:07am UTC](https://boards.straightdope.com/t/dumb-ass-algebra-question/230437/8 "2004-02-22T01:07:35Z")
> [@Achernar](#): > > You can’t break it into two equation like that. > > 1.1^x/1.2^x = (1.1/1.2)^x = (11/12)^x > > Now, as **g8rguy** says, to solve (11/12)^x = 5/7, you need to use logarithms. If you don’t remember how to work with these, you’ll have to remind yourself. > > log(11/12)^x = log(5/7) > x log(11/12) = log(5/7) > x = log(5/7) / log(11/12) = 3.8670 > > If you’re more comfortable saying, instead, x = log[sub]11/12/sub, that’s the same thing. I prefer the first way, though. OK, I just spent two _minutes_ working through it myself and proving conclusively that I’m a total dumbass. :o Yes, I was TOTALLY wrong and henceforth swear to never answer a math question I haven’t really looked at. (ARGH, my cheeks are literally burning. I shall slink away in shame…) ---
**Author:** ![Achernar](https://avatars.discourse-cdn.com/v4/letter/a/e274bd/32.png) [@Achernar](https://boards.straightdope.com/u/Achernar)\ **Post date:** [February 22, 2004, 1:09am UTC](https://boards.straightdope.com/t/dumb-ass-algebra-question/230437/9 "2004-02-22T01:09:14Z")
> [@Q.E.D.](#): > > I plugged the equation into MathCad, out of curiosity, and got the same result as **Achernar**. It came out to 3.8669912862235766523 (is it actually rational, or did MathCad round it off?) It’s not rational. I rounded too. ---
**Author:** ![gregongie](https://avatars.discourse-cdn.com/v4/letter/g/919ad9/32.png) [@gregongie](https://boards.straightdope.com/u/gregongie)\ **Post date:** [February 22, 2004, 1:09am UTC](https://boards.straightdope.com/t/dumb-ass-algebra-question/230437/10 "2004-02-22T01:09:22Z")
Argh. That final step should be: x = ln(5/7)/ln (1.1/1.2) Looks like others beat me to it anyway. You can use Base 10 Logs (log) or Base e Natural Logs (ln). You’ll get the same answer. ---
**Author:** ![bump](https://avatars.discourse-cdn.com/v4/letter/b/7c8e57/32.png) [@bump](https://boards.straightdope.com/u/bump)\ **Post date:** [February 22, 2004, 2:18am UTC](https://boards.straightdope.com/t/dumb-ass-algebra-question/230437/11 "2004-02-22T02:18:13Z")
Thanks guys! I thought it might be something like that, but I don’t have an algebra book anymore, and don’t even know anyone who does. ---
**Author:** ![RM\_Mentock](https://avatars.discourse-cdn.com/v4/letter/r/e274bd/32.png) [@RM\_Mentock](https://boards.straightdope.com/u/RM_Mentock)\ **Post date:** [February 22, 2004, 6:09am UTC](https://boards.straightdope.com/t/dumb-ass-algebra-question/230437/12 "2004-02-22T06:09:37Z")
> [@Q.E.D.](#): > > It came out to 3.8669912862235766523 (is it actually rational, or did MathCad round it off?) It’s transcendental.