Solve the given equation that involves fractional indices

[chwala]

Homework Statement:

This is my own created problem;

Solve for $x$ given

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

Relevant Equations:

understanding of indices

The actual problem that i was looking at with my students was supposed to be

$x^\frac{2}{3} - x^\frac{-2}{3}-6=0$(which is easy to solve using quadratic equations) of which i wanted them to solve, ...then i realized then that i had erronously posted

$x^\frac{2}{3} - x^\frac{-3}{2}-6=0$ on the board...and they were not able to proceed...

...anyway, i want to see if we can solve the problem as it is...


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

My take;

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

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

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

​

$\dfrac{2}{3}\log x=\log6$​

​

$\dfrac{2}{3}\log x=0.778$​

​

$\log x=1.167$​

​

$x=10^{1.167}=14.689$ (approximate solution in my opinion)​

​

My calculator indicates the solution as $x=14.7617$​

​

Your input is welcome...​

 

[SammyS]

Homework Statement:: This is my own created problem;

Solve for $x$ given

$x^\frac{2}{3} - x^\frac{-3}{2}-6=0$
Relevant Equations:: understanding of indices

The actual problem that i was looking at with my students was supposed to be

$x^\frac{2}{3} - x^\frac{-2}{3}-6=0$(which is easy to solve using quadratic equations) of which i wanted them to solve, ...then i realized then that i had erronously posted

$x^\frac{2}{3} - x^\frac{-3}{2}-6=0$ on the board...and they were not able to proceed...

...anyway, i want to see if we can solve the problem as it is...


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

My take;

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

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

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

​

$\dfrac{2}{3}\log x=\log6$​

​

$\dfrac{2}{3}\log x=0.778$​

​

$\log x=1.167$​

​

$x=10^{1.167}=14.689$ (approximate solution in my opinion)​

​

My calculator indicates the solution as $x=14.7617$​

​

Your input is welcome...​

This looks like a "Magical Step". Please explain how this is "Legal".

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

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

​

You do know that in general: $\displaystyle \quad \log(A+B) \ne \log A + \log B$ .
Right?

 

[Mark44]

To elaborate on what @SammyS said, $\log(a + b) \ne \log(a) + \log(b)$.

 

[chwala]

but wait a minute,

$\log(6x^{1.5})=\log 6 +\log x^{1.5}$ right?

and that is what i used...unless i am not getting your query right.

i have

$\dfrac{13}{6}\log x - \log 6-1.5\log x-\log1=0$ in my working.

 

[Frabjous]

In your solution you have the equation
$\frac 2 3 log(x) = log(6)$ which is equivalent to $x^\frac 2 3 - 6 = 0$ which is different from your original equation.

The problem that was pointed out by @SammyS and @Mark44 is going from

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

to

$\frac {13} 6 log(x) -1.5log(x)= log(6)$

it looks like you are using

$log (x^\frac {13} 6 - 6 x^\frac 3 2)= log (x^\frac {13} 6) - log(6 x^\frac 3 2)$ which is incorrect.

 

[chwala]

Now i have the equation;

$36x^3-x^{\frac{26}{6}}+2x^{\frac{13}{6}}-1=0$

looks like i will have to explore on a suitable numerical method; to arrive at this solutions;

$x=0.289$ (which does not satisfy our original equation)

and

$x=14.762$

 

[SammyS]

Now i have the equation;

$36x^3-x^{\frac{26}{6}}+2x^{\frac{13}{6}}-1=0$

looks like i will have to explore on a suitable numerical method; to arrive at this solutions;

$x=0.289$ (which does not satisfy our original equation)

and

$x=14.762$

Where did this problem come from?

If this is from the same problem as in the Original Post., you need to explain your steps.

Please, No big leaps !

 

[chwala]

Ok i have;

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

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

$\dfrac{(x^{\frac{13}{6}} -1)(x^{\frac{13}{6}} -1)}{x^3} =36$

$x^\frac{13}{3} -2x^{\frac{13}{6}} +1=36x^3$

$x^\frac{26}{6} -2x^{\frac{13}{6}} +1=36x^3$

$36x^3-x^{\frac{26}{6}}+2x^{\frac{13}{6}}-1=0$

 

[Frabjous]

Now i have the equation;

$36x^3-x^{\frac{26}{6}}+2x^{\frac{13}{6}}-1=0$

Why do you feel that this is a superior version of your initial equation?

 

[chwala]

I haven't mentioned anywhere that it's superior rather I am trying to go around the problem, ofcourse it may take time...but I will for sure get somewhere...still thinking over it...if you have a better suggestion then I would appreciate.

...I know at the end I would have to use a suitable numerical method to arrive at the solution.

 

[julian]

Write $y = x^{\frac{1}{6}}$.

 

[SammyS]

Ok i have;

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

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

So this is equivalent to $\displaystyle \quad\quad x^{{13/6}} -1=6x^{{3/2}}$,
which you essentially had in the OP.
You then squared this equation and rearranged it to get the following.

$36x^3-x^{\frac{26}{6}}+2x^{\frac{13}{6}}-1=0$

So now you have an equation with an additional term. Not surprisingly, it's no easier to solve, plus it has an extraneous solution.

Either way, you still have to use a numerical or graphical method to solve it.

Using @julian's idea, let $u = x^{(1/6)}$.

Then $\displaystyle \quad\quad x^{{13/6}} -1=6x^{{3/2}}$

becomes: $\displaystyle \quad\quad u^{13} -1-6u^9=0$.

Maybe no easier to solve, but it looks less intimidating.

 

[Charles Link]

Rewriting as $u^9(u^4-6)=1$ we see there will be a solution for $u$ very near $u_o=6^{1/4}$.
Writing $u=u_o+\Delta$, you can then do a binomial expansion/Taylor series to get a good estimate of $\Delta$.
I get $\Delta \approx \frac{ 1}{864}$.
which then gives
$x =u^6 \approx 14.76$.

Edit: I see there should also be a solution near $u=-u_o$, for which I'll try to post a $\Delta$ momentarily. Yes, and I think once again $\Delta \approx \frac{ 1}{864}$. Computing $x$ and seeing the negative sign gets cancelled,I think this may be an extraneous solution, but I need to study it further.

 

[julian]

Wolfram gets 14.7618

 

[chwala]

Wolfram gets 14.7618

Thanks @julian , actually if you go through my posts...i was able to find the same solution using graphical method (i.e desmos graphing application to be specific).

I am interested in exploring a numerical method. I am looking at this ...

 

[Charles Link]

I am interested in exploring a numerical method. I am looking at this ...

Please see post 13. I think that got the $x=14.76$ rather easily.

 

[chwala]

Please see post 13. I think that got the $x=14.76$ rather easily.

Thanks @Charles Link ,I actually intended to check on your approach...I had already seen it...i want to carefully analyse it then react. Cheers mate...

 

[Charles Link]

You could also do an iterative method: $u=(6+\frac{1}{u^9})^{1/4}$, starting with $1/u^9=0$.
6^{.25}=1.5650846 Next (6+.01775)^.25=1.566241 Next (6+.017655)^.25=1.566235
Then computing $x$, $x$=1.566235^6=14.7619
Each time you put the new $u$ it into $1/u^9$ on the right side, etc.
Only a couple of cycles are needed and you start to get some very high precision.

 

[chwala]

You could also do an iterative method: $u=(6+\frac{1}{u^9})^{1/4}$, starting with $1/u^9=0$.
6^{.25}=1.5650846 Next (6+.01775)^.25=1.566241 Next (6+.017655)^.25=1.566235
Then computing $x$, $x$=1.566235^6=14.7619
Each time you put the new $u$ it into $1/u^9$ on the right side, etc.
Only a couple of cycles are needed and you start to get some very high precision.

@Charles Link thanks mate...actually I am a great fun of iterative techniques...let me look at this in a few hrs time...there was one I learnt in my undergraduate studies...so called Runge-Kutta method...

 

[chwala]

Write $y = x^{\frac{1}{6}}$.

This is brilliant! Man!

 

[chwala]

You could also do an iterative method: $u=(6+\frac{1}{u^9})^{1/4}$, starting with $1/u^9=0$.
6^{.25}=1.5650846 Next (6+.01775)^.25=1.566241 Next (6+.017655)^.25=1.566235
Then computing $x$, $x$=1.566235^6=14.7619
Each time you put the new $u$ it into $1/u^9$ on the right side, etc.
Only a couple of cycles are needed and you start to get some very high precision.

Interesting approach...just a quick question, can we take $\dfrac{1}{u^9}=0$? or are you assuming (supposedly, by using limits) that $u$ tends to increase rapidly towards infinity, thereby having the limit of the term approaching $0$?

 

[Mark44]

You could also do an iterative method: $u=(6+\frac{1}{u^9})^{1/4}$, starting with $1/u^9=0$.

I get what you're trying to say here, but taken literally, this doesn't make sense. $1/u^9 = 0$ doesn't have any solutions.

A better way to write this would be the recursive definition $u_{n+1} = (6 + \frac{1}{(u_n)^9})^{1/4}$, where $u_0 = 0$.

 

[SammyS]

I get what you're trying to say here, but taken literally, this doesn't make sense. $1/u^9 = 0$ doesn't have any solutions.

A better way to write this would be the recursive definition $u_{n+1} = (6 + \frac{1}{(u_n)^9})^{1/4}$, where $u_0 = 0$.

Might be better to start with $u_0=6^{1/4}$ .

 

[Mark44]

Might be better to start with $u_0=6^{1/4}$ .

That would be my $u_1$.

 

[Charles Link]

$u_o=0$ gives $+\infty$ for $1/u_o^9$.

I also made that same mistake, thinking $u_o$ should be zero, but I edited it=post 18, before too many people saw it. (I think @chwala might have seen the unedited post=he gave me a "like" right after I edited it).

Meanwhile how to handle it formally, as @chwala asks in post 21, is open to debate. The $1/u_o^9$ is a $\Delta$ type term. What @SammyS proposes (post 23) is one way that may work ok, but yes, as @Mark44 says, $u_1$ should be $6^{1/4}$...

 

[Mark44]

On reflection, I concur with @SammyS's suggestion to start with $u_1 = 6^{1/4}$ rather than $u_0 = 0$. I didn't think through the ramifications of my earlier suggestion.

I'm fairly sure that you could start with $u_0 = 1$ and get a sequence that converges quickly.

 

[Mark44]

Starting with $u_0 = 1$ and using Excel, I get $u_4=1.566233106$ and the same values (to that precision) for subsequent terms in the sequence. $(u_4)^6 = 14.76176876$.

 

[Charles Link]

I think it may be worthwhile to go back to post 1 and see why an algebraic error resulted in a result that was close, but still incorrect. The result we get for $x$ is that $x \approx 14.7$. We have $x^{13/6}-1=6x^{3/2}$. The algebraic error ignored the minus $1$. If we put in $14.7$ for $x$ on the left side and ignore the minus $1$, we get $14.7^{13/6}= 338=6 x^{3/2}$, so that $x \approx (338/6)^{2/3}=14.695$.
Meanwhile with the minus $1$ , the $338$ becomes $337$, and we get $(337/6)^{2/3}=14.666$. (We see the approximate effect of ignoring the minus $1$.)
So we see how it gets close to the right answer, but is in any case incorrect.

Edit: Perhaps a better demonstration comes from using $x \approx 6^{3/2}$ (see post 5) which is what we have if we ignore the minus $1$. You can then do a Taylor series with $x=6^{3/2}+\Delta$, (just with first order terms, where $f(x)=x^{13/6}$ and $g(x)=6x^{3/2}$, expanding about $x_o=6^{3/2}$, etc.), where then with the minus $1$, we get $\Delta=1/(4 \cdot 6^{3/4})$. (Note $\Delta=0$ without the minus $1$.)

and note this $x \approx 6^{3/2}$ is the same thing as the $u \approx 6^{1/4}$ that we have above (posts 18-25) in the iterative approach. Ignoring the minus $1$ is the same thing as ignoring the $1/u^9$ term in the iterative approach.

It is interesting that @chwala 's algebraic error in the OP turns out to be of the $\Delta$ variety.
In general, $\log(a+b) \neq \log{ a}+\log{b}$.
In this case $\ln(a+b)=\ln(a(1+\frac{b}{a}))=\ln{a}+\ln(1+\frac{b}{a}) \approx \ln{a}+\frac{b}{a}$ for small $\frac{b}{a}$, and that explains why it got a result as close as it did.

 

[SammyS]

I think it may be worthwhile to go back to post 1 and see why an algebraic error resulted in a result that was close, but still incorrect. The result we get for $x$ is that $x \approx 14.7$. We have $x^{13/6}-1=6x^{3/2}$. The algebraic error ignored the minus $1$. If we put in $14.7$ for $x$ on the left side and ignore the minus $1$, we get $14.7^{13/6}= 338=6 x^{3/2}$, so that $x \approx (338/6)^{2/3}=14.695$.

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:

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

​

but wait a minute,

$\log(6x^{1.5})=\log 6 +\log x^{1.5}$ right?

and that is what i used...unless i am not getting your query right.

i have

$\dfrac{13}{6}\log x - \log 6-1.5\log x-\log1=0$ in my working.

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.

 

[Charles Link]

@SammyS It really turned into an interesting math problem that @chwala has, and with your preferring to work with the $x$ instead of $u$, (where $x=u^6$), the iterative formula is
$x=(6+\frac{1}{x^{3/2}})^{3/2}$.
Perhaps this is a little simpler than introducing a couple of large exponents, and then converting back to $x=u^6$.
It takes just about 2 or 3 cycles of this formula to get several decimal place accuracy, starting with $x=6^{3/2}$ on the right side.
In just two cycles, the answer I get agrees with @Mark44 's post 27 to 5 decimal places, and with one or two more cycles, complete agreement. :)
Edit: and I didn't even need Excel=all I needed to do was put the expression in the google search bar e.g.(6+1/(14.7622^(3/2)))^(3/2) and insert the new result by simply changing the 22 in the 14.7622 etc. and repeating the search.

 

[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.