# Math question **URL:** \ **Category:** Factual Questions\ **Created:** [April 9, 2011, 5:44pm UTC](https://boards.straightdope.com/t/math-question/577787 "2011-04-09T17:44:53Z")\ **Posts on this page:** 9\ **Page:** 1
**Author:** ![Johnny\_L.A](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/johnny_l.a/32/1084_2.png) [@Johnny\_L.A](https://boards.straightdope.com/u/Johnny_L.A)\ **Post date:** [April 9, 2011, 5:44pm UTC](https://boards.straightdope.com/t/math-question/577787/1 "2011-04-09T17:44:53Z")
The following appeared in _The Seattle Times_ [reformatted for the message board]: 1. Write (1 - 3_i_) ÷ (-2 - 4_i_) in the form a + b_i_. Simple algebra, I know. But it’s been a while. I haven’t worked on it, but at a glance I see that a 2 can be factored out of the denominator. I could multiply either the numerator or the denominator by its compliment, creating a perfect square. I’m otherwise occupied at the moment and don’t have the time to play with it right now, so I thought I’d post here to A) see if I’m on the right track; and B) to let others join in the fun. The answer is: 1/2 + (1/2)_i_ There are two other questions. The first one is solvable at a glance. The second (which is the third question) I haven’t tried yet. 1. Solve for x: 8[sup]x[/sup] = 32 As I said, it’s solvable at a glance: 5/3 1. The graph of the equation y = 3[sup]x[/sup] is reflected over the y-axis. What is the equation of the image? y = (1/3)[sup]x[/sup] ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [April 9, 2011, 5:55pm UTC](https://boards.straightdope.com/t/math-question/577787/2 "2011-04-09T17:55:22Z")
Everything you’ve said is correct. (Except “The second (which is the third question) I haven’t tried yet”. Seems to me you’ve answered all the questions…) ---
**Author:** ![Indistinguishable](https://avatars.discourse-cdn.com/v4/letter/i/90ced4/32.png) [@Indistinguishable](https://boards.straightdope.com/u/Indistinguishable)\ **Post date:** [April 9, 2011, 5:58pm UTC](https://boards.straightdope.com/t/math-question/577787/3 "2011-04-09T17:58:09Z")
Oh, wait, I see. You’ve read the answers and are posting them all, but don’t know how to obtain them. Heh, sorry, I misinterpreted you. ---
**Author:** ![MikeS](https://avatars.discourse-cdn.com/v4/letter/m/919ad9/32.png) [@MikeS](https://boards.straightdope.com/u/MikeS)\ **Post date:** [April 9, 2011, 6:00pm UTC](https://boards.straightdope.com/t/math-question/577787/4 "2011-04-09T18:00:32Z")
> [@Johnny\_L.A](#): > > The following appeared in _The Seattle Times_ [reformatted for the message board]: > > 1. Write (1 - 3_i_) ÷ (-2 - 4_i_) in the form a + b_i_. > > Simple algebra, I know. But it’s been a while. I haven’t worked on it, but at a glance I see that a 2 can be factored out of the denominator. I could multiply either the numerator or the denominator by its compliment, creating a perfect square. I’m otherwise occupied at the moment and don’t have the time to play with it right now, so I thought I’d post here to A) see if I’m on the right track; and B) to let others join in the fun. I think you mean “conjugate” rather than “compliment”, but other than that, yes. The standard MO for finding the real and imaginary part of (a + bi)/(c + di) is to multiply top and bottom by (c - di). The denominator becomes real, and you can multiply out the numerator to get the real and imaginary parts. ---
**Author:** ![emacknight](https://avatars.discourse-cdn.com/v4/letter/e/3ec8ea/32.png) [@emacknight](https://boards.straightdope.com/u/emacknight)\ **Post date:** [April 9, 2011, 6:03pm UTC](https://boards.straightdope.com/t/math-question/577787/5 "2011-04-09T18:03:01Z")
Question [1] isn’t simple algebra. i is an imaginary number squareroot(-1). But the solution is simple in that you [spoiler]multiply the top and bottom by the conjugate of the bottom -2+4i where i \* i = -1 (1 - 3i) \* (-2 +4i) = -2 +4i +6i +12 = 10 +10i (-2 -4i) \* (-2 +4i) = 4 -8i + 8i +16 = 20 1/2 + (1/2)i[/spoiler] ---
**Author:** ![panamajack](https://avatars.discourse-cdn.com/v4/letter/p/47e85d/32.png) [@panamajack](https://boards.straightdope.com/u/panamajack)\ **Post date:** [April 9, 2011, 6:08pm UTC](https://boards.straightdope.com/t/math-question/577787/6 "2011-04-09T18:08:04Z")
The third is probably the easiest - if you’re reflecting across the y axis, for any value of the function at x, the reflection will have the value at -x. Thus for r the reflection, r(x) = f(-x). ---
**Author:** ![Johnny\_L.A](https://sea3.discourse-cdn.com/straightdope/user_avatar/boards.straightdope.com/johnny_l.a/32/1084_2.png) [@Johnny\_L.A](https://boards.straightdope.com/u/Johnny_L.A)\ **Post date:** [April 9, 2011, 6:11pm UTC](https://boards.straightdope.com/t/math-question/577787/7 "2011-04-09T18:11:24Z")
> [@MikeS](#): > > I think you mean “conjugate” rather than “compliment”, but other than that, yes. The standard MO for finding the real and imaginary part of (a + bi)/(c + di) is to multiply top and bottom by (c - di). The denominator becomes real, and you can multiply out the numerator to get the real and imaginary parts. ‘Conjugate’. I said it’s been a while. :smack: That’s what I thought. I’m still elsewhere, but I’ll work through it after a while. **emacknight** : Yes, I assumed people would know that _i_ is the square root of negative one. ---
**Author:** ![Wendell\_Wagner](https://avatars.discourse-cdn.com/v4/letter/w/8491ac/32.png) [@Wendell\_Wagner](https://boards.straightdope.com/u/Wendell_Wagner)\ **Post date:** [April 10, 2011, 3:14am UTC](https://boards.straightdope.com/t/math-question/577787/8 "2011-04-10T03:14:35Z")
Incidentally, this appeared in an article written originally for _The Washington Post_ about whether it would be a good idea to require Algebra II for all high school students: > **[Requiring Algebra II in high school gains momentum nationwide](https://www.washingtonpost.com/business/economy/requiring-algebra-ii-in-high-school-gains-momentum-nationwide/2011/04/01/AF7FBWXC_story.html)** > > A national push has more states requiring Algebra II of high school graduates. But critics say: What’s the point? ---
**Author:** ![SimonMoon5](https://avatars.discourse-cdn.com/v4/letter/s/9f8e36/32.png) [@SimonMoon5](https://boards.straightdope.com/u/SimonMoon5)\ **Post date:** [April 10, 2011, 3:45pm UTC](https://boards.straightdope.com/t/math-question/577787/9 "2011-04-10T15:45:56Z")
When I teach people how to work that first problem, I tell them that we’re doing the same trick we do when we try to rationalize a denominator that contains two terms (which at worst contain square roots). Because the problem in both sorts of problems is that you have a square root in the denominator that you don’t want there; with a complex number, the square root you’re trying to get rid of is the square root of -1 (aka _i_). So, you just do the same thing. Trying to write (1 - 3_i_) ÷ (-2 - 4_i_) in the form a + b_i_ involves what I call “real-izing” the denominator, by which I mean changing the denominator into a real number. That’s no harder than trying to simply (or rationalize the denominator of) (1 - 3_sqrt(2)) ÷ (-2 - 4_sqrt(2)), where I’m using sqrt(2) to represent the square root of 2. In both problems, you multiply the numerator and denominator by the conjugate of the denominator. So, anybody who can successfullly rationalize denominators shouldn’t have much trouble with dividing a complex number by a complex number to get a number written in standard complex number form. * * * Now, for the 8^x = 32 problem, what we teach in algebra classes is that if a \> 0 and a \<\>1, then a^b = a^c if and only if b = c. So, it’s just a matter of writing both sides of the equation as 2 raised to a power and then equating the powers. Thus: since 8 = 2^3 and 32 = 2^5, we have (2^3)^x = 2^5, or in other words 2^3x = 2^5, so 3x = 5, so x = 5/3. * * * The work for the other problem has already been described.