what's wrong here?

[allegansveritatem]

Here is the problem that I am supposed to solve without a calculator:



Here is what I did:



Seems to me something has taken a left turn here but...what? I have a feeling I am headed here for a log with an argument of 0, which is a breach of mathematical propriety.

 

[lev888]

Are you sure about -2?

 

[Dr.Peterson]

Also, there is no rule that says that if ab = 1, then a=1 and b=1. That's only true if the RHS is 0, not 1. How about a=2 and b=1/2?

Instead, combine the factors into a single power ...

 

[Deleted member 4993]

Here is the problem that I am supposed to solve without a calculator:

View attachment 15060

Here is what I did:

View attachment 15061

Seems to me something has taken a left turn here but...what? I have a feeling I am headed here for a log with an argument of 0, which is a breach of mathematical propriety.

I would do the problem following way:

log(x)^(1/2) = [log(x)]^(1/2)

let log(x) = u then

1/2 * u = u^(1/2)

u^(1/2) * [1/2*u^(1/2) - 1] = 0

one of the solutions: u^(1/2) = 0

The other one:

1/2*u^(1/2) - 1= 0

u^(1/2) = 2

u = 4 and continue.......

 

[JeffM]

I had some initial trouble following Subhotosh Khan's excellent but somewhat abbreviated suggestion.

[MATH]\text {Let } u = log_b(x).[/MATH]
[MATH]\therefore \sqrt{log_b(x)} = \sqrt{u} = u^{(1/2)}.[/MATH] Straightforward, right?

[MATH]log_b(\sqrt{x}) = log_b(x^{(1/2)}) = \dfrac{1}{2} * log(x) = \dfrac{u}{2}.[/MATH] Again straightforward.

[MATH]\therefore \sqrt{log_b(x)} = log_b(\sqrt{x}) \implies u^{(1/2)} = \dfrac{u}{2} \implies 2\sqrt{u} = u \implies[/MATH]
[MATH]4u = u^2 \implies u = 0 \text { or } u = 4.[/MATH]
I have followed SK's suggestion, but filled in some details. Finish it up.

 

[pka]

View attachment 15060 Seems to me something has taken a left turn here but...what? I have a feeling I am headed here for a log with an argument of 0, which is a breach of mathematical propriety.

Without a calculator:
Rewrite as \(\displaystyle \dfrac{1}{2}\log(x)=\sqrt{\log(x)}\) Now let \(\displaystyle u=\log(x)\)
Then squaring we get
\(\displaystyle u^2=4u\\u^2-4u=0\\u(u-4)=0\\u=0\text{ or }u=4\)
That means \(\displaystyle x=e^0 = 1\text{ or }x=e^{4}\)...........................................edited

 

[Steven G]

You knew that x^-2 = 1/x² and thought that sqrt(x) = 1/²

 

[Deleted member 4993]

You knew that x^-2 = 1/x² and thought that sqrt(x) = 1/²

Jomo, Which post are you responding to?

What is that?

 

[allegansveritatem]

Are you sure about -2?

I am ashamed to say that I used -2 when I should have used 1/2. I don't know what I was thinking about.

 

[allegansveritatem]

Also, there is no rule that says that if ab = 1, then a=1 and b=1. That's only true if the RHS is 0, not 1. How about a=2 and b=1/2?

Instead, combine the factors into a single power ...

right. I think my mind was on holiday yesterday. I knew when I wrote that bit about 1 being the only thing etc., that I was writing poetry and passing it off for math logic.

 

[allegansveritatem]

I would do the problem following way:

log(x^)(1/2) = [log(x)]^(1/2)

let log(x) = u then

1/2 * u = u^(1/2)

u^(1/2) * [1/2*u^(1/2) - 1] = 0

one of the solutions: u^(1/2) = 0

The other one:

1/2*u^(1/2) - 1= 0

u^(1/2) = 2

u = 4 and continue.......

I see it. Substitution. I'll try it. Thanks

 

[allegansveritatem]

I had some initial trouble following Subhotosh Khan's excellent but somewhat abbreviated suggestion.

[MATH]\text {Let } u = log_b(x).[/MATH]
[MATH]\therefore \sqrt{log_b(x)} = \sqrt{u} = u^{(1/2)}.[/MATH] Straightforward, right?

[MATH]log_b(\sqrt{x}) = log_b(x^{(1/2)}) = \dfrac{1}{2} * log(x) = \dfrac{u}{2}.[/MATH] Again straightforward.

[MATH]\therefore \sqrt{log_b(x)} = log_b(\sqrt{x}) \implies u^{(1/2)} = \dfrac{u}{2} \implies 2\sqrt{u} = u \implies[/MATH]
[MATH]4u = u^2 \implies u = 0 \text { or } u = 4.[/MATH]
I have followed SK's suggestion, but filled in some details. Finish it up.

will do. Thanks

 

[allegansveritatem]

Without a calculator:
Rewrite as \(\displaystyle \dfrac{1}{2}\log(x)=\sqrt{\log(x)}\) Now let \(\displaystyle u=\log(x)\)
Then squaring we get
\(\displaystyle u^2=4u\\u^2-4u=0\\u(u-4)=0\\u=0\text{ or }u=4\)
That means \(\displaystyle x=e^0 = 1\text{ or }x=e^{4}\)...........................................edited

yes, I think I know the way to go now. Thanks

 

[allegansveritatem]

You knew that x^-2 = 1/x² and thought that sqrt(x) = 1/²

yes, I know that. Somehow yesterday, however, I stopped knowing it long enough to screw up this problem royally.

 

[allegansveritatem]

I went at it again today, long and strong, and here is what I came up with--in two tries. I feel I am coming closer but am somehow missing something:



and



I worked these out keeping in mind what was said in the replies to my original post but not in the presence of those replies because I wanted to solve the problem in my own way. But...I can't see where I am still going WRONG.

 

[pka]

I went at it again today, long and strong,I worked these out keeping in mind what was said in the replies to my original post but not in the presence of those replies because I wanted to solve the problem in my own way. But...I can't see where I am still going WRONG.

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 \(\displaystyle x=1~\&~x=e^4\).

 

[JeffM]

[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.

 

[Steven G]

Why do you have trouble with solutions by inspection?

Students are usually not taught to look for an obvious solution.

When I was a graduate student I went head to head with my professor over this and amazingly lost.
I handed in a Linear Algebra assignment which his TA graded. The TA marked one of my problems wrong because I showed no work, but stated as easily seen the answer is ... I went to my professor to complain and he said that he always goes with his TA' s decision. So I asked him if this meant that he did not want me to think anymore? He wouldn't respond. I was pissed off at this. A PhD professor not giving credit to a graduate student who thought before doing the mechanics to the problem. Somethings you just don't ever forget!

 

[pka]

Students are usually not taught to look for an obvious solution.

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:



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?"

 

[lookagain]

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 \(\displaystyle x=1~\&~x=e^4\).

No, it is neither "incompletely obvious" nor "completely obvious."