Important * means multiply (a) means base a etc
log(b)a*log(c)b*log(a)c=1
Hello,Originally Posted by kingkaisai2
as you may know you can transform a logarithm, so you have only one base:
$\displaystyle \log_{b}{(a)}={\ln(a)\over\ln{(b)}$
Your problem can be written like this:
$\displaystyle {\ln(a)\over\ln{(b)}} \cdot {\ln(b)\over\ln{(c)}} \cdot {\ln(c)\over\ln{(a)}} =1$
Simplify the LHS of this equation and you'll get 1 = 1. Thus this equation is true for all $\displaystyle \{a, b, c\}\in\mbox{]0;1[} \cup \mbox{ ]1;\infty[}$
(Remark: The $\displaystyle \cup$ means union of 2 sets of numbers. )
Greetings
EB
Hello, kingkaisai2!
If you don't know the Base-Change Formula,$\displaystyle \log_b(a)\cdot\log_c(b)\cdot\log_a(c)\:=\:1$
. . you'll have to do some Olympic-level gymnastics.
Let $\displaystyle \log_c(b) = P$ . . . then: $\displaystyle c^P = b$
Take logs (base b): $\displaystyle \log_b(c^P) = \log_b(b)\quad\Rightarrow\quad P\cdot\log_b(c) = 1$
Then: $\displaystyle P = \frac{1}{\log_b(c)} $ . . . That is: $\displaystyle \log_c(b) = \frac{1}{\log_b(c)} $
Let $\displaystyle \log_a(c) = Q$ . . . then: $\displaystyle a^Q = c$
Take logs (base b): $\displaystyle \log_b(a^Q) = \log_b(c)\quad\Rightarrow\quad Q\cdot\log_b(a) = \log_b(c)$
Then: $\displaystyle Q = \frac{\log_b(c)}{\log_b(a)}$ . . . That is: $\displaystyle \log_a(c) = \frac{\log_b(c)}{\log_b(a)}$
So: $\displaystyle \log_b(a)\cdot\log_c(b)\cdot\log_a(c)$ becomes:
. . $\displaystyle \log_b(a)\cdot\frac{1}{\log_b(c)}\cdot\frac{\log_b (c)}{\log_b(a)} \;=\;1 $ . . . ta-DAA!