Can x be solved in this logarithmic equation?

[Dr-NiKoN]

Say you have:
$\frac{\log(x)}{r\log{x}} = y$

r and y can be any given number.

Is there any way to solve this for x?

ie

$\frac{\log(x)}{1.6\log{x}} = 20$

I'm getting this because I'm trying to calculate $f(x) = \frac{x}{a}$ where f(x) is of the form $f(x) = cx^r$
$a > 1$

 

[marlon]

Say you have:
$\frac{\log(x)}{r\log{x}} = y$

r and y can be any given number.

Is there any way to solve this for x?

ie

$\frac{\log(x)}{1.6\log{x}} = 20$

I'm getting this because I'm trying to calculate $f(x) = \frac{x}{a}$ where f(x) is of the form $f(x) = cx^r$
$a > 1$

I don't get this, sorry. correct me if i am wrong here but i have :

$$\frac {\log(x)}{r\log(x)} = y$$ but doesn't the left hand side yield
$$\frac {1}{r}$$ ?

i am missing the point here...you sure about this equation ?

marlon

 

[Dr-NiKoN]

Well, I'm trying to find x for $f(x) = \frac{x}{a}$

where f is of the form $cx^r$

I tried setting $x = cx^r$ and taking the log to find x.

I was wondering if it was possible to find x that way, or maybe if there is another way?

 

[marlon]

$cx^r = \frac {x}{a}$
$cx^r - \frac {x}{a} = 0$
$x * (cx^{r-1} - \frac {1}{a}) = 0$
$cx^{r-1} = \frac {1}{a}$ if x is not 0
$x^{r-1} = \frac {1}{ca}$ if x is not 0
$x = \sqrt [r - 1] {\frac {1}{ca}}$ if x is not 0


marlon

 

[marlon]

what do you say about that ?

marlon

 

[Dr-NiKoN]

Hm, there must be a more elegant way?

 

[marlon]

trust me , this is the most easy way out. working with log will make you go around in circles

marlon

 

[Dr-NiKoN]

Hehe, ok, works for me.

I'm finding myself confused looking for the more complex ways to solve stuff, when regular math will do the job. Oh well :)

thanks a lot.

 

[matt grime]

If you did your logs properly it would also drop out:

cx^r=x/a

then, assuming x and every thing else for that matter, is positive,

logc + rlogx = logx - loga

(r-1)logx= -loga - log c

log x = -log(ac) * 1/(r-1)

or x = (ac)^{1/(1-r)

exactly as marlon showed without needing to use logs.