20221026, 16:58  #1 
"Kyle"
Feb 2005
Somewhere near M52..
2·3^{3}·17 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
19×83 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
2^{6}×3^{2}×11 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..
2·3^{3}·17 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
2^{6}·3^{2}·11 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:


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