Math Help - Solution to equation

1. Solution to equation

I was in the shower and I was just thinking is the solution to $1-\log_a(x)=-\log_x(a)$

$x=a^{\phi}$?

2. Yop,

What is $\phi$?

3. Originally Posted by Moo
Yop,

What is $\phi$?
The golden ratio $\phi=\frac{1+\sqrt{5}}{2}$

4. Well, I find the same thing yep ~

I can't work with $\log_a$ :

$1-\frac{\ln(x)}{\ln(a)}=-\frac{\ln(a)}{\ln(x)}$

$\ln(a)\ln(x)-\ln^2(x)=-\ln^2(a)$

Dividing by $\ln^2(a)$ :

$(\frac{\ln(x)}{\ln(a)})^2-\frac{\ln(x)}{\ln(a)}-1=0$

Hence $x=a^\varphi$

Is it what you were thinking about ?

PS : I rather prefer $\varphi$ than $\phi$

5. Originally Posted by Moo
Well, I find the same thing yep ~

I can't work with $\log_a$ :

$1-\frac{\ln(x)}{\ln(a)}=-\frac{\ln(a)}{\ln(x)}$

$\ln(a)\ln(x)-\ln^2(x)=-\ln^2(a)$

Dividing by $\ln^2(a)$ :

$(\frac{\ln(x)}{\ln(a)})^2-\frac{\ln(x)}{\ln(a)}-1=0$

Hence $x=a^\varphi$

PS : I rather prefer $\varphi$ than $\phi$
Isnt that cool...I did it a different way $1-\log_a(x)=-\log_x(a)\Rightarrow{1-\log_a(x)=\frac{-1}{log_a(x)}}$...then letting $\log_a(x)=u$ and multiplying through and using quadratic formula I came up with $\phi$...thats right $\phi$

6. Originally Posted by Mathstud28
Isnt that cool...I did it a different way $1-\log_a(x)=-\log_x(a)\Rightarrow{1-\log_a(x)=\frac{-1}{log_a(x)}}$...then letting $\log_a(x)=u$ and multiplying through and using quadratic formula I came up with $\phi$...thats right $\phi$
this is the way i'd do it

7. Interesting..I wonder why I've never seen that before..

I'm just relieved to find out that I'm not the only one who thinks about math in the shower.

8. Originally Posted by elizsimca
Interesting..I wonder why I've never seen that before..
yes, it is interesting.

How did you come up with the question Mathstud?

I'm just relieved to find out that I'm not the only one who thinks about math in the shower.
nope, you're the only one. Mathstud was joking when he said that...