what's wrong here?

pka said:
PROBLEM: \(\displaystyle \log(\sqrt x)=\sqrt{\log(x)}\).
Why do you have trouble with solutions by inspection?
It is incompletely obvious that the solution is \(\displaystyle x=1~\&~x=e^4\).
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You say it is obvious but I will have to patiently work out the connections between these terms and the terms in the problem...not till then, I'm afraid, will it be obvious for me.
 
JeffM said:
[MATH]log(\sqrt{x}) = y = \sqrt{log(x)}.[/MATH] That is given.

[MATH]\therefore 10^y = \sqrt{x} \implies x = 10^{2y} = 10^{2\sqrt{log(x)}} \implies[/MATH]
[MATH]log(x) = 2\sqrt{log(x)}.[/MATH]
[MATH]\text {Let } u = log(x) \implies u = 2\sqrt{u} \implies u^2 = 4u.[/MATH]
So far so good. Now you go off track. You go

[MATH]u^2 = 4u \implies \dfrac{u^2}{u} = \dfrac{4u}{u}.[/MATH]
But this is not necessarily true. What is true is

[MATH]u \ne 0 \ \& \ u^2 = 4u \implies \dfrac{u^2}{u} = \dfrac{4u}{u} \implies u = 4 \implies[/MATH]
[MATH]log(x) = 4 \implies x = 10000.[/MATH]
Is x = 10000 a valid solution?

[MATH]log(\sqrt{10000}) = log(100) = 2.[/MATH]
[MATH]\sqrt{log(10000)} = \sqrt{4} = 2.[/MATH]
Somehow you made a mistake when checking.

So x = 10000 is a valid solution. Moreover, it is the only valid solution if u does not equal 0.

[MATH]u = 0 \implies log(x) = 0 \implies x = 1.[/MATH]
[MATH]log(\sqrt{1}) = log(1) = 0.[/MATH]
[MATH]\sqrt{log(1)} = \sqrt{0} = 0.[/MATH]
Where did you get [MATH]u^2 = u[/MATH] You had 4u.
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I can't find where I have U squared = U....I think you are referring to one of the equations where I had put a 2 on the RS and then, realizing it was a mistake, wrote a 4 over the two so that neither the 2 nor the 4 was clear and thus it gave the effect of a crossing out rather than a correction. But...what is flummoxing me in your reply is this equation: U squared/U = 4U/U. I mean, where does the U in the RS denominator come from? Isn't it just U squared/U=4U?
 
allegansveritatem said:
You say it is obvious but I will have to patiently work out the connections between these terms and the terms in the problem...not till then, I'm afraid, will it be obvious for me.
Click to expand...

It is not obvious to me, nor is that a proper way to solve an algebra problem by itself. It can, however, be part of solving by inspection (together with proof that there can be no other solution). In this case, I don't think that's appropriate.
 
pka said:
I love to tell this real story. In the third grade my daughter's teacher handed out a test containing a pie chart of eight equal parts. The instructions were to color in half the pie . Well my daughter turned in the following graph:

View attachment 15084

It was marked incorrect and sent home for review and parental signature.
As luck would have it the teacher was currently in a graduate mathematics education course I was holding.
I simply wrote a cryptic note: "Since when is four eights not one half?"
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good story....not all teachers are thinkers.
 
allegansveritatem said:
I can't find where I have U squared = U....I think you are referring to one of the equations where I had put a 2 on the RS and then, realizing it was a mistake, wrote a 4 over the two so that neither the 2 nor the 4 was clear and thus it gave the effect of a crossing out rather than a correction. But...what is flummoxing me in your reply is this equation: U squared/U = 4U/U. I mean, where does the U in the RS denominator come from? Isn't it just U squared/U=4U?
Click to expand...

JeffM's comments were about things you wrote in post #15.

In the first image, you had [MATH]u^2 = 4u \implies \dfrac{u^2}{u} = 4[/MATH]. You divided both sides by [MATH]u[/MATH], though you didn't explicitly state what you were doing.

The division is correct; the problem is that you shouldn't ever divide by a variable, because you don't know if it is zero. In doing so, you lost the fact that [MATH]u = 0[/MATH] is also a solution. Factoring is the preferred way, but you can also separately check whether 0 would yield a solution.

In the second image, you had [MATH]u^2 = u[/MATH], which seems to come from nowhere. (You had dropped the radical on the line before.)
 
allegansveritatem said:
I can't find where I have U squared = U....I think you are referring to one of the equations where I had put a 2 on the RS and then, realizing it was a mistake, wrote a 4 over the two so that neither the 2 nor the 4 was clear and thus it gave the effect of a crossing out rather than a correction. But...what is flummoxing me in your reply is this equation: U squared/U = 4U/U. I mean, where does the U in the RS denominator come from? Isn't it just U squared/U=4U?
Click to expand...
In post 15, you asked where you had gone wrong in your next attempt. I answered but did not refer to that post. Sorry for the confusion. You made three mistakes. I discussed them all.
 
Dr.Peterson said:
JeffM's comments were about things you wrote in post #15.

In the first image, you had [MATH]u^2 = 4u \implies \dfrac{u^2}{u} = 4[/MATH]. You divided both sides by [MATH]u[/MATH], though you didn't explicitly state what you were doing.

The division is correct; the problem is that you shouldn't ever divide by a variable, because you don't know if it is zero. In doing so, you lost the fact that [MATH]u = 0[/MATH] is also a solution. Factoring is the preferred way, but you can also separately check whether 0 would yield a solution.

In the second image, you had [MATH]u^2 = u[/MATH], which seems to come from nowhere. (You had dropped the radical on the line before.)
Click to expand...

It didn't occur to me that when dividing by a variable I was excluding zero, but that is certainly what it means. thanks for pointing this out.
Yes, in the second image I squared both sides, and eventually came up with 1 as a (possible) solution. And this seems, as a possibility, correct. When I did the problem another way, I got 10,000 as a possible solution, which, it turns out, is the only right answer here. So, does that mean the solution in the second image is wholly false? But, I guess, since I divided by a variable, that vitiates the whole business, even if the answer I came up with is technically a plausibility.
 
so, here is what I did today, before I learned that dividing by a variable is not good practice. So, I thought to make this the final shot but it is not to be. I will have to go back and do it again.
aaaaaaaaatenthou.PNG

I post this with what intent I can't say. It is not done right, although it is the right answer. Sometimes things work out that way. It's called, I believe, dumb luck.
 
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aaaaaaaaaten thousand.PNG

Reading this over, I see that I have not finished. There is the little matter of the square of log x = log x^4.... I am going back to the drawing board.
 
allegansveritatem said:
I have discovered tonight that I don't know how to do this without dividing by the variable.
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Posts 5 and 6 demonstrate how to do this: rather than divide x^2 = 4x by x, collect terms on one side and factor out the x: x^2 - 4x = 0 becomes x(x - 4) = 0, which can be factored easily. This is an important general technique; I tell my students that division is a natural temptation in such problems, and factoring is the valid alternative - the way to escape the temptation.
 
allegansveritatem said:
Yes, in the second image I squared both sides, and eventually came up with 1 as a (possible) solution. And this seems, as a possibility, correct. When I did the problem another way, I got 10,000 as a possible solution, which, it turns out, is the only right answer here. So, does that mean the solution in the second image is wholly false? But, I guess, since I divided by a variable, that vitiates the whole business, even if the answer I came up with is technically a plausibility.
Click to expand...
No, in that second image in post #15, you went from [MATH](\log\sqrt{x})^2 = \log x[/MATH] to [MATH](\log x)^2 = \log x[/MATH], losing the radical on the LHS apparently by mere forgetfulness -- nothing you did would cause it to go away. So this is not a valid solution. It was wrong before you divided by u.

Do you still think it is correct?
 
Dr.Peterson said:
No, in that second image in post #15, you went from [MATH](\log\sqrt{x})^2 = \log x[/MATH] to [MATH](\log x)^2 = \log x[/MATH], losing the radical on the LHS apparently by mere forgetfulness -- nothing you did would cause it to go away. So this is not a valid solution. It was wrong before you divided by u.

Do you still think it is correct?
Click to expand...
No, it can't be correct. What I seem to have done is to square the square root of log X--the right side of the original equation--twice. I squared both sides and got log square root of x in parentheses squared for LHS and log x on RHS which is a result of squaring square root of log x. Then I....well as I study it I can't quite tell what I did. But yes, it seems that I did forget the radical....at any rate something got squared twice.
 
Dr.Peterson said:
No, in that second image in post #15, you went from [MATH](\log\sqrt{x})^2 = \log x[/MATH] to [MATH](\log x)^2 = \log x[/MATH], losing the radical on the LHS apparently by mere forgetfulness -- nothing you did would cause it to go away. So this is not a valid solution. It was wrong before you divided by u.

Do you still think it is correct?
Click to expand...
I see it. I knew one of the above posts had the problem worked out but I wanted first to see if I could get it on my own. I took the suggestion re substitution as salvational but more was needed. I will go the factoring route and post results, which I feel will be impeccable. Thanks for taking the time to check my work.
 
You get to [MATH]u^2 = 4u.[/MATH]
You now have at least three ways to solve this equation.

ONE: QUADRATIC FORMULA

[MATH]u^2 = 4u \implies u^2 - 4u + 0 = 0 \implies u = \dfrac{4 \pm \sqrt{16 - 4 * 1 * 0}}{2 * 1} \implies[/MATH]
[MATH]u = \dfrac{4 \pm \sqrt{16}}{2} = \dfrac{4 \pm 4}{2} \implies[/MATH]
[MATH]u = 0 \text { or } u = 4.[/MATH]
TWO: BY CASES

Without reference to the equation, it is true that

[MATH]u = 0 \text { or } u \ne 0.[/MATH]
With reference to the equation, Is it true that

[MATH]0^2 = 4 * 0 \implies 0 * 0 = 0 \implies 0 = 0.[/MATH] Yes, it's true.

[MATH]u \ne 0 \text { and } u^2 = 4u \implies \dfrac{u^2}{u} = \dfrac{4u}{u} \implies u = 4.[/MATH]
[MATH]\therefore u = 0 \text { or } u = 4.[/MATH]
THREE: FACTORING AND THE ZERO PRODUCT PROPERTY

[MATH]u^2 = 4u \implies u^2 - 4u = 0 \implies u(u - 4) = 0 \implies[/MATH]
[MATH]u = 0 \text { or } u = 4 \text { by the zero product property.}[/MATH]
Of the three methods, the quadratic formula is narrowest because it applies only to quadratics. The factoring method applies to all polynomials without a constant term, is the quickest, and is the least likely to lead to error,. Solution by cases is very general, but is time consuming and more likely to lead to error because you must cover each case.

So the factoring method is the recommended approach.
 
JeffM said:
You get to [MATH]u^2 = 4u.[/MATH]
You now have at least three ways to solve this equation.

ONE: QUADRATIC FORMULA

[MATH]u^2 = 4u \implies u^2 - 4u + 0 = 0 \implies u = \dfrac{4 \pm \sqrt{16 - 4 * 1 * 0}}{2 * 1} \implies[/MATH]
[MATH]u = \dfrac{4 \pm \sqrt{16}}{2} = \dfrac{4 \pm 4}{2} \implies[/MATH]
[MATH]u = 0 \text { or } u = 4.[/MATH]
TWO: BY CASES

Without reference to the equation, it is true that

[MATH]u = 0 \text { or } u \ne 0.[/MATH]
With reference to the equation, Is it true that

[MATH]0^2 = 4 * 0 \implies 0 * 0 = 0 \implies 0 = 0.[/MATH] Yes, it's true.

[MATH]u \ne 0 \text { and } u^2 = 4u \implies \dfrac{u^2}{u} = \dfrac{4u}{u} \implies u = 4.[/MATH]
[MATH]\therefore u = 0 \text { or } u = 4.[/MATH]
THREE: FACTORING AND THE ZERO PRODUCT PROPERTY

[MATH]u^2 = 4u \implies u^2 - 4u = 0 \implies u(u - 4) = 0 \implies[/MATH]
[MATH]u = 0 \text { or } u = 4 \text { by the zero product property.}[/MATH]
Of the three methods, the quadratic formula is narrowest because it applies only to quadratics. The factoring method applies to all polynomials without a constant term, is the quickest, and is the least likely to lead to error,. Solution by cases is very general, but is time consuming and more likely to lead to error because you must cover each case.

So the factoring method is the recommended approach.
Click to expand...
I will have to think about the middle method but I have used the first and the last. Factoring is nice...but it is applicable only when the numbers are fairly small or at least fairly neat. I mean, 2 and 4, 20 and 5 etc. or 100 and 1000 etc.
 
So, I think I have it now and I find that both 1 and 10000 are solutions for x and 0 and 4 for log x. Please excuse the sloppy correction in the last lines. I had to write with a mouse, something that is not so easy to do.

logx square2.PNG
 
The work is good. But the checks look as if you were thinking incorrectly, improperly simplifying each side.

To check a solution, do not do any of the work you do in solving (such as squaring each side); just evaluate each side separately. The check for x = 10,000 looks like this:

[MATH]\log\sqrt{10,000} = \log{100} = 2[/MATH]​
[MATH]\sqrt{\log{10,000}} = \sqrt{4} = 2[/MATH]​

Since these numbers are equal, x = 10,000 is a solution.

This is important because it is too easy to make the same mistakes you might have made when you solved; evaluating is safer, and at the least would involve different mistakes.
 
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