mersenneforum.org Help with logarithms containing negative arguments
 User Name Remember Me? Password
 Register FAQ Search Today's Posts Mark Forums Read

 2022-10-26, 16:58 #1 Primeinator     "Kyle" Feb 2005 Somewhere near M52.. 7·131 Posts Help with logarithms containing negative arguments Hey, all. I've been out of school for school for some number of years now, about 13 years since I took ODE. I came a across a fun question on social media. 4^x + 6^x = 9^x (for real solution of x) which I was able to solve without too much difficulty though took me longer than it should have, perhaps. The imaginary solution will come from: (log ((1 - sqr(5))/2) / log (3/2)) Converting to natural log and expanding out: (Ln e^(i*pi)) + ln (-1 + sqr(5)) - ln (2)) / ln (3/2) Using only the principal log for the first term I get: (i*pi + ln (-1 + sqr(5)) - ln (2)) / ln (3/2) This is very cumbersome. Is there a simpler/more elegant way to write this, assuming it is even correct? I couldn't think of another way to handle the negative argument other than factoring out the negative one. Thanks! And my sincerest apologies for any errors made.
 2022-10-26, 20:50 #2 lycorn     "GIMFS" Sep 2002 Oeiras, Portugal 1,571 Posts I tried it and found exactly the same solutions. Logarithms with negative arguments are complex numbers, "cumbersome" as it may look. Last fiddled with by lycorn on 2022-10-26 at 20:53
 2022-10-27, 13:42 #3 Dr Sardonicus     Feb 2017 Nowhere 3×31×67 Posts Let's see here. The OP's answer can be obtained as follows: 4^x + 6^x = 9^x Divide through by 4^x 1 + (3/2)^x = (9/4)^x Let (3/2)^x = a. Then 1 + a = a^2 a^2 - a - 1 = 0 This quadratic equation has two roots, $\varepsilon\;>\;1\text{ and }\frac{-1}{\varepsilon}$. Using the positive root we get $x\;=\;\frac{\log(\varepsilon)}{\log$$\frac{3}{2}$$}$. Using the real-valued logarithms, all is well. Using the negative root, we have $$$\frac{3}{2}$$^{x}\;=\;\frac{-1}{\varepsilon}$. Taking log(-1/u) = log(-1) - log(u) where u positive and real, and log(u) is real, we get the posted answer, with log(-1) = pi*i. You can also use any odd integer multiple of pi*i for log(-1). Note that 4^x is defined as exp(log(4)*x) and similarly for the other exponentials. It is important that you use the real-valued logarithms of 4, 6, 9, 3/2, and 9/4. Otherwise, the "laws of exponents" will come to grief.
2022-10-27, 14:21   #4
Primeinator

"Kyle"
Feb 2005
Somewhere near M52..

7×131 Posts

Quote:
 Originally Posted by lycorn I tried it and found exactly the same solutions. Logarithms with negative arguments are complex numbers, "cumbersome" as it may look.
I suppose "cumbersome" was not the most appropriate word. I just want it to look more elegant but instead it looks like last year's attempt at pumpkin carving.

Quote:
 Originally Posted by Dr Sardonicus Taking log(-1/u) = log(-1) - log(u) where u positive and real, and log(u) is real, we get the posted answer, with log(-1) = pi*i. You can also use any odd integer multiple of pi*i for log(-1). Note that 4^x is defined as exp(log(4)*x) and similarly for the other exponentials. It is important that you use the real-valued logarithms of 4, 6, 9, 3/2, and 9/4. Otherwise, the "laws of exponents" will come to grief.
This is my first attempt at a negative logarithms, actually. If I recall, our professor in calculus said, "It is not technically true you cannot take the log of a negative number but we are not going to cover that now". I watched a brief YouTube video that noted you could multiply i*pi by any integer (given it will just be a rotation of 180 degrees on the unit circle on the complex plane). Is there a reason you listed only positive integers in this example, or is this more a matter of semantics?

"Otherwise the laws of exponents will come to grief". This is rather poetic. Are you saying that even though I can manipulate log (1 - sqr(5)) to log (-1) + log(-1 + sqr(5)), or similar examples, that it will not satisfy the original equation?

2022-10-27, 16:30   #5
Dr Sardonicus

Feb 2017
Nowhere

3·31·67 Posts

Quote:
 Originally Posted by Primeinator Is there a reason you listed only positive integers in this example, or is this more a matter of semantics?
I didn't. I said you could use any odd integer multiple of p*i as a logarithm of -1.

Quote:
 "Otherwise the laws of exponents will come to grief". This is rather poetic. Are you saying that even though I can manipulate log (1 - sqr(5)) to log (-1) + log(-1 + sqr(5)), or similar examples, that it will not satisfy the original equation?
No, in your problem the choice of "x" isn't going to cause trouble. As I said, it's using the real logarithms of 4, 6, 3/2 and 9/4 that is important.

As a simple example of the kind of thing that can go wrong with the "laws of exponents," note that (-1)*(-1) = +1. Now, raise both sides to the 1/2 power:

((-1)*(-1))^(1/2) = 1^(1/2). If you try to apply the "law of exponents" that (x*y)^e = x^e*y^e you get (-1)^(1/2)*(-1)^(1/2) = 1^(1/2).

That is, sqrt(-1)*sqrt(-1) = +1. Uh-oh, trouble!

What went wrong? Raising a number to non-integer powers depends on its logarithm, that's what. And "the" logarithm is only determined up to an integer multiple of 2*pi*i.

Take any value of log(-1), say pi*i. Then the log of -1*-1 = 1 is log(-1) + log(-1) = 2*pi*i. But when we say 1^(1/2) = 1, we are implicitly using the value 0 for log(1).

It is only by using the real logarithms of 4, 3/2, 6 and 9/4 that you can be sure that 4^x*(3/2)^x = 6^x and (9/4)^x = ((3/2)^x)^2. Fortunately the value of x being "the" logarithm of a negative number isn't going to cause any trouble. What's important is that log(4) + log(3/2) = log(6) and 2*log(3/2) = log(9/4).

2022-10-28, 03:37   #6
sweety439

"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36

2·7·263 Posts

Quote:
 Originally Posted by Primeinator Hey, all. I've been out of school for school for some number of years now, about 13 years since I took ODE. I came a across a fun question on social media. 4^x + 6^x = 9^x (for real solution of x) which I was able to solve without too much difficulty though took me longer than it should have, perhaps. The imaginary solution will come from: (log ((1 - sqr(5))/2) / log (3/2)) Converting to natural log and expanding out: (Ln e^(i*pi)) + ln (-1 + sqr(5)) - ln (2)) / ln (3/2) Using only the principal log for the first term I get: (i*pi + ln (-1 + sqr(5)) - ln (2)) / ln (3/2) This is very cumbersome. Is there a simpler/more elegant way to write this, assuming it is even correct? I couldn't think of another way to handle the negative argument other than factoring out the negative one. Thanks! And my sincerest apologies for any errors made.
Logarithms with negative arguments are complex numbers, e.g. log(-1) = pi*i, 3*pi*i, 5*pi*i, 7*pi*i, ... (it has infinitely many possible values, i.e. (2*k+1)*pi*i with (positive or negative) integer k)

 Similar Threads Thread Thread Starter Forum Replies Last Post mart_r Math 2 2019-03-21 22:03 hydeer Miscellaneous Math 4 2018-05-28 03:42 Raman Factoring 1 2016-05-23 13:44 Lakshmi Factoring 10 2013-12-16 20:27 bsquared Programming 12 2008-12-05 03:22

All times are UTC. The time now is 00:20.

Wed Feb 8 00:20:24 UTC 2023 up 173 days, 21:48, 1 user, load averages: 1.43, 1.21, 1.17

Copyright ©2000 - 2023, Jelsoft Enterprises Ltd.

This forum has received and complied with 0 (zero) government requests for information.

Permission is granted to copy, distribute and/or modify this document under the terms of the GNU Free Documentation License, Version 1.2 or any later version published by the Free Software Foundation.
A copy of the license is included in the FAQ.

≠ ± ∓ ÷ × · − √ ‰ ⊗ ⊕ ⊖ ⊘ ⊙ ≤ ≥ ≦ ≧ ≨ ≩ ≺ ≻ ≼ ≽ ⊏ ⊐ ⊑ ⊒ ² ³ °
∠ ∟ ° ≅ ~ ‖ ⟂ ⫛
≡ ≜ ≈ ∝ ∞ ≪ ≫ ⌊⌋ ⌈⌉ ∘ ∏ ∐ ∑ ∧ ∨ ∩ ∪ ⨀ ⊕ ⊗ 𝖕 𝖖 𝖗 ⊲ ⊳
∅ ∖ ∁ ↦ ↣ ∩ ∪ ⊆ ⊂ ⊄ ⊊ ⊇ ⊃ ⊅ ⊋ ⊖ ∈ ∉ ∋ ∌ ℕ ℤ ℚ ℝ ℂ ℵ ℶ ℷ ℸ 𝓟
¬ ∨ ∧ ⊕ → ← ⇒ ⇐ ⇔ ∀ ∃ ∄ ∴ ∵ ⊤ ⊥ ⊢ ⊨ ⫤ ⊣ … ⋯ ⋮ ⋰ ⋱
∫ ∬ ∭ ∮ ∯ ∰ ∇ ∆ δ ∂ ℱ ℒ ℓ
𝛢𝛼 𝛣𝛽 𝛤𝛾 𝛥𝛿 𝛦𝜀𝜖 𝛧𝜁 𝛨𝜂 𝛩𝜃𝜗 𝛪𝜄 𝛫𝜅 𝛬𝜆 𝛭𝜇 𝛮𝜈 𝛯𝜉 𝛰𝜊 𝛱𝜋 𝛲𝜌 𝛴𝜎𝜍 𝛵𝜏 𝛶𝜐 𝛷𝜙𝜑 𝛸𝜒 𝛹𝜓 𝛺𝜔