Logarithm problem

  • #1

Homework Statement



Given 3log5 xy2 + log5x = 4+2log5y, prove xy = 5.

Homework Equations





The Attempt at a Solution



Can i prove like this? Or i must use 3log5 xy2 + log5x = 4+2log5y to prove xy=5? Meaning my final result must be xy = 5? Please enlighten me..

Mathematics 3.png
 

Answers and Replies

  • #2
ehild
Homework Helper
15,543
1,914
It is a correct proof if you add the condition that both x and y are greater then 0, just to be strict.

ehild
 
  • #3
22,129
3,298
I don't get it. You use xy=5, to prove that relationship.

But you must do the opposite. You must use that relationship to prove xy=5. That means that you can't start from assuming xy=5. You must prove it.
 
  • #4
Curious3141
Homework Helper
2,850
87
Technically, you've proven that xy = 5 satisfies the equation (that xy = 5 is sufficient). However, I believe the question is asking you to prove that xy = 5 is the only solution (that xy = 5 is necessary).

You'd be better advised to just manipulate the equation with the laws of logs until you get the required result. It's quite simple.
 
  • #5
Curious3141
Homework Helper
2,850
87
I don't get it. You use xy=5, to prove that relationship.

But you must do the opposite. You must use that relationship to prove xy=5. That means that you can't start from assuming xy=5. You must prove it.

Beat me to the punch. :tongue:
 
  • #6
ehild
Homework Helper
15,543
1,914
I have problems with the meaning of the sentence "prove xy=5" Can not it be understood as "prove xy=5 is solution of the equation"? When we solve an equation, we substitute the result back to see if it is really a solution.

I think the other procedure, starting from the original equation would correspond to the task: "prove that xy=5 follows from equation
3 log5 xy2 + log5 x = 4+2 log5 y ".

ehild
 
  • #7
103
1
Technically,

you've proven that xy = 5 satisfies the equation
(that xy = 5 is sufficient).
[itex]\text{If xy = 5 is known to be true, then it}[/itex]
[itex]\text{is necessary as part (or all) of the answer.}[/itex]



However, I believe the question is asking you to prove that xy = 5 is the only solution
(that xy = 5 is necessary).

[itex]\text{If xy = 5 is the only solution, then that would be sufficient, because it would suffice.}[/itex]

.............
 
  • #8
Curious3141
Homework Helper
2,850
87
I have problems with the meaning of the sentence "prove xy=5" Can not it be understood as "prove xy=5 is solution of the equation"? When we solve an equation, we substitute the result back to see if it is really a solution.

I think the other procedure, starting from the original equation would correspond to the task: "prove that xy=5 follows from equation
3 log5 xy2 + log5 x = 4+2 log5 y ".

ehild

The way the question is phrased, viz. *Given* that equation (call this proposition [itex]P[/itex]), *prove* xy=5 (call this proposition [itex]Q[/itex]) indicates that one must establish [itex]P \Rightarrow Q[/itex].

Substituting xy = 5 back into the equation only establishes [itex]Q \Rightarrow P[/itex]. I don't think this is sufficient.

And, just to be pedantic, when one solves an equation, one tries to find all solutions (or at least all solutions within a defined range) that satisfy an equation. Substituting values back into an equation to see that LHS = RHS is essentially verifying that those values are solutions, not solving the equation. :smile:
 
  • #9
Curious3141
Homework Helper
2,850
87
.............

Sorry, your post is unclear, and I can't even tell if you're agreeing with me. But see my reply to ehild's post.
 
  • #10
103
1
Technically, you've proven that xy = 5
satisfies the equation (that xy = 5 is sufficient). However,
I believe the question is asking you to prove that xy = 5 is
the only solution (that xy = 5 is necessary).

Here is my take:


To prove that xy = 5 is the only solution, does it suffice to show
that xy = 5 ? No, but it is necessary that xy = 5.

Suppose it is shown that xy = 5 satisfies the equation. It would
be sufficient to show that no other solution satisfies the equation
in order to prove that xy = 5 is the only solution.
 
  • #11
Curious3141
Homework Helper
2,850
87
Here is my take:


To prove that xy = 5 is the only solution, does it suffice to show
that xy = 5 ? No, but it is necessary that xy = 5.

Suppose it is shown that xy = 5 satisfies the equation. It would
be sufficient to show that no other solution satisfies the equation
in order to prove that xy = 5 is the only solution.

OK, so how would you show there are no other solutions?
 
  • #12
Curious3141
Homework Helper
2,850
87
To illustrate my point, if I posed a question like this:

Given [itex]\log_5{x} + \log_5{y} = \frac{1}{\log_5{x} + \log_5{y}}[/itex] prove that [itex]xy = 5[/itex],

it would actually be an incorrect question. Because it doesn't allow for the possibility that [itex]xy = \frac{1}{5}[/itex], also a valid solution.

However, if I posed the question like this:

Verify that [itex]xy = 5[/itex] satisfies the equation [itex]\log_5{x} + \log_5{y} = \frac{1}{\log_5{x} + \log_5{y}}[/itex],

that would be perfectly alright and the OP's method would be completely valid.

It's not the best example, but it's the best argument I could come up with while suffering from a bad cold. :tongue:
 
  • #13
103
1
OK, so how would you show there are no other solutions?

After using logarithmic and expontial rules, one of the steps
I got to is:

[itex](xy)^4 = 5^4[/itex]


I see that xy = 5 works (with the acceptable ranges of x and y).



Then I would show that all of the remaining cases

xy = -5, xy = 5i, and xy = -5i fail to satisfy the equation,
when taking into account the permissible real values for x and y.




Note: I left this forum for the night.
 
  • #14
Curious3141
Homework Helper
2,850
87
After using logarithmic and expontial rules, one of the steps
I got to is:

[itex](xy)^4 = 5^4[/itex]


I see that xy = 5 works (with the acceptable ranges of x and y).



Then I would show that all of the remaining cases

xy = -5, xy = 5i, and xy = -5i fail to satisfy the equation,
when taking into account the permissible real values for x and y.




Note: I left this forum for the night.

OK, I suppose that's fair enough. I guess any rigorous method one uses to solve this problem would involve observing that x and y should be restricted to positive values. It's probably easier to just state that at the start, as ehild suggests instead of seeking then excluding imaginary solutions after the fact.
 
Last edited:
  • #15
ehild
Homework Helper
15,543
1,914
And, just to be pedantic, when one solves an equation, one tries to find all solutions (or at least all solutions within a defined range) that satisfy an equation. Substituting values back into an equation to see that LHS = RHS is essentially verifying that those values are solutions, not solving the equation. :smile:

Thanks Curious. I know that you are right, I just quibble ... :smile: But it is usual in high-school Maths when showing that two relations P an G are equivalent that we say: we got Q from P through equivalence relations, so the proof works backwards, too. Applying the same relations in the opposite order we would get also Q→P.

To admit the truth, it is not the case here. xy=5 follows from the equation supposed it is valid, that is, both x and y are positive, but not vice versa. The last step of the proof log x +log y →log(xy), is true, but log(xy)→log x+log y is not without excluding non-positive x, y values.

To show that there is only one solution of the original equation it is enough to refer that it is linear in the logarithms and logarithm is a monotonous function:lg(x)=log(a)→x=a Proving that xy=5 is solution means also that it is the only solution.


ehild
 
Last edited:
  • #16
785
15

Homework Statement



Given 3log5 xy2 + log5x = 4+2log5y, prove xy = 5.

Homework Equations





The Attempt at a Solution



Can i prove like this? Or i must use 3log5 xy2 + log5x = 4+2log5y to prove xy=5? Meaning my final result must be xy = 5? Please enlighten me..

View attachment 42605

This problem is not difficult :

Just solve 3log5 xy2 + log5x = 4+2log5y using law : log a + log b = log ab

You will surely get your way out easily rather than doing opposite...
See post #2 also.
 
  • #17
Curious3141
Homework Helper
2,850
87
Thanks Curious. I know that you are right, I just quibble ... :smile: But it is usual in high-school Maths when showing that two relations P an G are equivalent that we say: we got Q from P through equivalence relations, so the proof works backwards, too. Applying the same relations in the opposite order we would get also Q→P.

To admit the truth, it is not the case here. xy=5 follows from the equation supposed it is valid, that is, both x and y are positive, but not vice versa. The last step of the proof log x +log y →log(xy), is true, but log(xy)→log x+log y is not without excluding non-positive x, y values.

To show that there is only one solution of the original equation it is enough to refer that it is linear in the logarithms and logarithm is a monotonous function:lg(x)=log(a)→x=a Proving that xy=5 is solution means also that it is the only solution.


ehild

Good point. Actually, if one wishes to consider logs of negative numbers, both ways can be iffy, depending on how one defines the log. Considering only principal values of the natural logs, [itex]\ln(-1) = i\pi[/itex] and [itex]\ln1 = 0[/itex] (the latter as usual).

Even [itex]\log x + \log y = \log{xy}[/itex] doesn't work out as [itex]2\ln(-1) = 2{\pi}i \neq 0 = \ln1[/itex]. And going the other way (splitting it up) similarly doesn't work if we want to write [itex]0 = \ln1 \neq 2{\pi}i = 2\ln(-1)[/itex].

I understand this, and I guess I was tacitly assuming that we were working with positive x,y (since this is at the elementary level). But it's better to state this explicitly, as you suggested. And the "resubstituting" (verifying) sort of solution is fine if (as you suggest), it's made rigorous by observing that logs are one-to-one.
 

Related Threads on Logarithm problem

  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
12
Views
5K
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
4
Views
739
  • Last Post
Replies
6
Views
3K
Replies
13
Views
1K
  • Last Post
Replies
12
Views
3K
  • Last Post
Replies
10
Views
2K
Top