Can x be solved in this logarithmic equation?

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Discussion Overview

The discussion revolves around solving a logarithmic equation of the form \(\frac{\log(x)}{r\log{x}} = y\) for \(x\), with specific examples provided. Participants explore different approaches to find \(x\) in the context of a function \(f(x) = \frac{x}{a}\) where \(f\) is expressed as \(cx^r\). The conversation includes attempts to clarify the equation and various methods for solving it.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants propose that the equation \(\frac{\log(x)}{r\log{x}} = y\) can be manipulated to find \(x\), while others express confusion about the validity of the equation.
  • One participant suggests setting \(x = cx^r\) and taking the logarithm to find \(x\), questioning if this method is viable.
  • Another participant derives an expression \(cx^r - \frac{x}{a} = 0\) and simplifies it to find \(x\), indicating a potential solution without using logarithms.
  • There is a suggestion that working with logarithms may complicate the problem unnecessarily, with one participant advocating for a more straightforward algebraic approach.
  • A later reply confirms that a proper manipulation of logarithms leads to a similar result as the algebraic method proposed earlier.

Areas of Agreement / Disagreement

Participants express differing views on the best method to solve for \(x\), with some advocating for logarithmic approaches and others favoring algebraic methods. The discussion does not reach a consensus on a single preferred method.

Contextual Notes

There are indications of confusion regarding the manipulation of logarithmic expressions, and assumptions about the positivity of \(x\) and other variables are made without explicit confirmation. The discussion reflects varying levels of comfort with logarithmic versus algebraic methods.

Dr-NiKoN
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Say you have:
[itex]\frac{\log(x)}{r\log{x}} = y[/itex]

r and y can be any given number.

Is there any way to solve this for x?

ie

[itex]\frac{\log(x)}{1.6\log{x}} = 20[/itex]

I'm getting this because I'm trying to calculate [itex]f(x) = \frac{x}{a}[/itex] where f(x) is of the form [itex]f(x) = cx^r[/itex]
[itex]a > 1[/itex]
 
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Dr-NiKoN said:
Say you have:
[itex]\frac{\log(x)}{r\log{x}} = y[/itex]

r and y can be any given number.

Is there any way to solve this for x?

ie

[itex]\frac{\log(x)}{1.6\log{x}} = 20[/itex]

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

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

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

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

marlon
 
Well, I'm trying to find x for [itex]f(x) = \frac{x}{a}[/itex]

where f is of the form [itex]cx^r[/itex]

I tried setting [itex]x = cx^r[/itex] 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?
 
[itex]cx^r = \frac {x}{a}[/itex]
[itex]cx^r - \frac {x}{a} = 0[/itex]
[itex]x * (cx^{r-1} - \frac {1}{a}) = 0[/itex]
[itex]cx^{r-1} = \frac {1}{a}[/itex] if x is not 0
[itex]x^{r-1} = \frac {1}{ca}[/itex] if x is not 0
[itex]x = \sqrt [r - 1] {\frac {1}{ca}}[/itex] if x is not 0


marlon
 
what do you say about that ?

marlon
 
Hm, there must be a more elegant way?
 
trust me , this is the most easy way out. working with log will make you go around in circles

marlon
 
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.
 
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.
 

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