      # More Solutions May Exist

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### #1 squonk

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Posted 27 September 2008 - 03:07 PM

I'm trying the CP Manager and I've solved this equation with solve function:

log(2,x^2)+(log(2,x))^2=0

The CP give me this message "More solutions may exist" and give the result:

{x=1}

I know that the equation has another solution:

{x=1/4}

Why this second solution is not shown? What does exactly means the message?
On the CP manual I've found this:

For the solution, the solve function returns an expression or value for the expression (Exp/Eq) input as its argument. The message "More solutions may exist" will appear on the display when a value is returned as the solution, because there may be multiple solutions.

The CP can not calculate all the solutions? Can someone explain?
Many thanks!

### #2 Guest_Nemo_*

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Posted 28 September 2008 - 02:31 PM

Assume you want to solve this:
sin(x)=0.5

how many solutions may exist?

x = pi/6
x = pi - pi/6
.
.
.

Am I clear?

### #3 Kilburn

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Posted 28 September 2008 - 03:39 PM

I may be retarded, but since when are log functions periodic?

### #4 kucalc

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Posted 28 September 2008 - 04:55 PM

Same thing with the TI-nSpire, it only finds x = 1 and gives the warning "More solutions may exist".

I may be retarded, but since when are log functions periodic?

They aren't, but I think the guy is trying to make the point that you should do your own work.

### #5 squonk

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Posted 28 September 2008 - 08:55 PM

Assume you want to solve this:
sin(x)=0.5

how many solutions may exist?

x = pi/6
x = pi - pi/6

Am I clear?

I know what you mean, also in the CP manual is reported that if a equation has infinite solutions it report only 10...
But here I got an equation with 2 solutions (real numbers, not complex) and the CP report only one of them.
I conclude that the algorithm that CP use can not find all solutions.
It could be interesting to investigate why and how the algorithm is done...

For kucalc:
yes, I've also tried with TI-Nspire and also with Maxima, the same result.
With Derive I have found this:

x^2 * e ^(LN(x)^2 /LN(2)) = 1
???

I have tried also other logarithmic equation with "strange" solutions...
I don't know why...

### #6 Guest_Guest_mccoy_*_*

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Posted 29 September 2008 - 08:50 AM

upgrade to at least ver 3.02 and you'll get all the solutions!

### #7 McCoy

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Posted 29 September 2008 - 09:53 AM

I'm trying the CP Manager and I've solved this equation with solve function:

log(2,x^2)+(log(2,x))^2=0

The CP give me this message "More solutions may exist" and give the result:

{x=1}

I know that the equation has another solution:

{x=1/4}

Why this second solution is not shown? What does exactly means the message?
On the CP manual I've found this:

For the solution, the solve function returns an expression or value for the expression (Exp/Eq) input as its argument. The message "More solutions may exist" will appear on the display when a value is returned as the solution, because there may be multiple solutions.

The CP can not calculate all the solutions? Can someone explain?
Many thanks!

Hi mate, I know there's a problem with the way calculators handle this question, but there's a way of getting around it. First of all change the base to exp (use natural log) i.e (ln(x))^2/(ln(2))^2+ln(x^2)/ln(2). Now solve this for x.If it doesn't give you the two solutions that you want, then smash that damn calculator/cpm.And don't waste time trying to solve this question on ti nspire or ti 89 ...no matter what you do, they won't give you the right solns for this question....(only cp/cpm can solve it).By the way, it's a shame that calculators can't solve this simple logarithm and you may be right if you're mad at them...there's no better way you should have done this question other than what you've tried.For God's sake, how can someone know that you have to change the base to get the right answer? And why does anyone have to do that anyway? well .....So just to let you know other option, I've also used mathematicals table and got the two solutions(x,1/4).Mathematical tables are slow to work with but they are more reliable than cp or any other calculator.

### #8 weirleader

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Posted 29 September 2008 - 11:18 PM

I'm trying the CP Manager and I've solved this equation with solve function:

log(2,x^2)+(log(2,x))^2=0

The CP give me this message "More solutions may exist" and give the result:

{x=1}

What version of the OS are you using? What mode are you in? I couldn't recreate the problem on my CP - it gave me both solutions... :-)

### #9 squonk

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Posted 29 September 2008 - 11:54 PM

Thanks to all for replies!

upgrade to at least ver 3.02 and you'll get all the solutions!

I've downloaded the trial version from one of the Casio site. The trial version is 3.00.3510.2160 that could be the first 3.00 version.
I've tried to find update to the latest version, but you must be registered user to download the last version.
Can someone help me to find latest version of CP Manager?

Hi mate, I know there's a problem with the way calculators handle this question, but there's a way of getting around it. First of all change the base to exp (use natural log) i.e (ln(x))^2/(ln(2))^2+ln(x^2)/ln(2). Now solve this for x.If it doesn't give you the two solutions that you want, then smash that damn calculator/cpm.And don't waste time trying to solve this question on ti nspire or ti 89 ...no matter what you do, they won't give you the right solns for this question....(only cp/cpm can solve it).By the way, it's a shame that calculators can't solve this simple logarithm and you may be right if you're mad at them...there's no better way you should have done this question other than what you've tried.For God's sake, how can someone know that you have to change the base to get the right answer? And why does anyone have to do that anyway? well .....So just to let you know other option, I've also used mathematicals table and got the two solutions(x,1/4).Mathematical tables are slow to work with but they are more reliable than cp or any other calculator.

I've tried to solve the equation with changing base, but I've got the same result.
Is then a problem with 3.00 version?
Can you tell me why the TI calculators can not handle in the right way this type of equations?
And about mathematicals table: can you explain?

What version of the OS are you using? What mode are you in? I couldn't recreate the problem on my CP - it gave me both solutions... :-)

About OS I have already written: I want however to underline that I don't got the CP calculator, but I'am simply testing the CP Manager.
I've tried to solve the equation in Real Mode, but I think that the problem could be the version of mine CPM...

### #10 vanhoa

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Posted 30 September 2008 - 05:05 AM

OS 3.0: ( http://casio.vanhoav...09-08/solve.PNG )

OS 3.02: ( http://casio.vanhoav...9-08/solve2.PNG )

### #11 squonk

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Posted 30 September 2008 - 02:08 PM

Thanks vanhoa for this screenshots!
I don't know why in Real e Complex mode the result could be different: the algorithm to find the solution must be unveiled...
Also is a mistery why with the simple changing the equation the result comes both!

So I ask: can someone help me to find CP Manager ver 3.02 or 3.03?
Many thanks!

### #12 Guest_Nemo_*

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Posted 08 October 2008 - 04:09 PM

Thanks vanhoa for this screenshots!
I don't know why in Real e Complex mode the result could be different: the algorithm to find the solution must be unveiled...
Also is a mistery why with the simple changing the equation the result comes both!

So I ask: can someone help me to find CP Manager ver 3.02 or 3.03?
Many thanks!

excuse me for delay, I go here rarely.

As you know, In Complex view of Log function, we have a periodic function:

z1 = r e^(i*Theta)
z2 = r e^(i*(Theta + 2k Pi))

you see that z1 = z2, because:

z1 = r cos(Theta) + i * r sin(Theta)
z2 = r cos(Theta + 2k Pi) + i * r sin(Theta + 2k Pi)

which is the same as z1.

Now, take a Log of z1 and z2:
z1 = z2

Log(z1) = Log® + Log(e^(i*Theta)) = Log® + i * (Theta) * Log(e)

Log(z2) = Log® + Log(e^(i*Theta + 2k Pi)) = Log® + i * (Theta) * Log(e) + i * 2k Pi * Log(e)

You see that we have an imaginary part which makes a periodic behavior.

Summary:
1. Log is periodic in Complex point of view.
2. I think that CP does all calculation in Complex way, even when you choose Real. after computing all results, it does a conversion to real mode.

Am i clear this time?

### #13 Guest_Nemo_*

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Posted 08 October 2008 - 04:34 PM

I may be retarded, but since when are log functions periodic?

excuse for the second post!

My dear friend, Kilburn
may I ask you have a look at the following url:

`http://math.fullerton.edu/mathews/c2003/ComplexFunLogarithmMod.html`

you can see that in general we can not write : (e^(w))^z = e^(wz).

In general, when you change the form of an equation, its complex roots may change, so CP may be successful to find the new roots.

By the way, I have written a term improperly, take a look to the last post:

Log(z2) = LogÂ® + Log(e^(i*Theta + 2k Pi)).....

I should place parenthesis like this:
Log(z2) = LogÂ® + Log(e^(i*(Theta + 2k Pi))).....

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