20221026, 16:58  #1 
"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. 
20221026, 20:50  #2 
"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 20221026 at 20:53 
20221027, 13:42  #3 
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, . Using the positive root we get . Using the realvalued logarithms, all is well. Using the negative root, we have . 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 realvalued logarithms of 4, 6, 9, 3/2, and 9/4. Otherwise, the "laws of exponents" will come to grief. 
20221027, 14:21  #4  
"Kyle"
Feb 2005
Somewhere near M52..
7×131 Posts 
Quote:
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? 

20221027, 16:30  #5  
Feb 2017
Nowhere
3·31·67 Posts 
Quote:
Quote:
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. Uhoh, trouble! What went wrong? Raising a number to noninteger 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). 

20221028, 03:37  #6  
"99(4^34019)99 palind"
Nov 2016
(P^81993)SZ base 36
2·7·263 Posts 
Quote:


Thread Tools  
Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
sum (1/n^s), negative values of  mart_r  Math  2  20190321 22:03 
Using Logarithms of Mersenne Primes  hydeer  Miscellaneous Math  4  20180528 03:42 
Can discrete logarithms / factoring be done in P (i.e., deterministic polynomial time)?  Raman  Factoring  1  20160523 13:44 
Software for discrete logarithms  Lakshmi  Factoring  10  20131216 20:27 
arguments to main  bsquared  Programming  12  20081205 03:22 