Solve the given equation that involves fractional indices

[chwala]

I consider the OP error to be an error in applying the rules of logarithms rather than an algebraic error. It's pretty much sheer luck (IMO) that his approximate solution was so close to the correct value.

In Post #4, OP explains his thinking in going from:

$\displaystyle x^\frac{13}{6}-6x^\frac{3}{2}-1=0$

To:

$\displaystyle \dfrac{13}{6}\log x-1.5 \log x= \log 6$​

​

Not only was the log of a sum/difference handled incorrectly, but this implied that $\log(0)=0$.

Excuse my digression.

I chose to solve the equation as written in terms of $x$ rather than use the change variable to $u =x^{1/6}$. Consider the equation in essentially its given form:

$\displaystyle \quad\quad x^{{2/3}} -\dfrac 1 {x^{{3/2}}}=6$

Clearly the left hand side is dominated by $\displaystyle \quad x^{{2/3}}$.

So a good approximation is $\displaystyle x \approx 6^{3/2} \approx 14.697$ which is very close to what @chwala got. It's exactly what his approximate value would have been had he carried more digits in his intermediate steps.

Interesting @SammyS , can we then generalise by saying that one has to consider the dominant exponent when working out similar problems of this nature/kind then work out a recursive relationship to determine the approximate solution?your working step to solution looks pretty straightforward...

that is equations of the form,

$x^\frac{m}{n} - x^\frac{-n}{m}-k=0$ ...where $x^\frac{m}{n}$ is dominant. and $k$ is a constant.

 

[Charles Link]

@chwala
See
https://www.physicsforums.com/threads/the-solution-to-a-cubic-equation.975534/
for a problem you might find of interest that uses an iterative solution. I don't know whether there are any "rules" to getting a good iterative formula. Sometimes it helps just to be a little clever to spot something that will work.

 

[chwala]

@chwala
See
https://www.physicsforums.com/threads/the-solution-to-a-cubic-equation.975534/
for a problem you might find of interest that uses an iterative solution. I don't know whether there are any "rules" to getting a good iterative formula. Sometimes it helps just to be a little clever to spot something that will work.

I will look at this...

 

[malawi_glenn]

These particular indicies are called exponents

 

[chwala]

These particular indicies are called exponents

...sorry you meant 'indices' or is 'indicies' which is the the correct plural form? Sorry, English is not my first language... i always use the former in my lessons... i know that the singular form is index.

 

[Mark44]

'indices'

Indices -- the plural of index -- is the correct spelling. As far as I'm aware, there is no such word as indicies in English.

 

[chwala]

Indices -- the plural of index -- is the correct spelling. As far as I'm aware, there is no such word as indicies in English.

Thanks noted.

 

[epenguin]

It's not really English it's Greek. So not every native English speaker knows this particular singular-plural couple. (And actually that is true for a number of scientific terms.)

Oh, and while I am here, people have gone thoroughly through calculations of, I guess, some interest, but none of it had the point that your original equation before you mistranscribed it had.

 

[Mark44]

It's not really English it's Greek.

The word I said wasn't English was "indicies," a misspelling of indices. "Index" and its plural "indices" (also "indexes") are Latin in origin. See https://www.etymonline.com/word/index.

So not every native English speaker knows this particular singular-plural couple.

Absolutely. Some others include codex/codices, vortex/vortices, vertex/vertices, as well as words that end in -ix such as appendix/appendices, cicatrix/cicatrices (scar) and a few others. I suspect that these Latin plural endings will become obsolete if they aren't already.

 

[epenguin]

Exactly- some of us know, some of us don't know and some of us have a vague idea or feeling.