Need some help with solving a logarithmic equation

  • Thread starter Adir_Sh
  • Start date
  • #1
22
0

Homework Statement


The equation is as follows:
23w3z0l.gif


Homework Equations


What technique shall I use to solve the equation?

The Attempt at a Solution


Tried to solve it in any way that went through my mind, including making a substitution of terms but couldn't quite get the solution or find a path towards it.
 

Answers and Replies

  • #3
epenguin
Homework Helper
Gold Member
3,837
866
Start by rewriting the RHS without a √ symbol but only indices.
 
  • #4
34,697
6,399
Yes, show your work. At this site we require homework posters to show what they have tried. It's debatable that what you have in your post counts as a valid attempt.
 
  • #5
22
0
Here's some of the work I made. The problem for me is that "x" is found on both the base and the power of the exponent, and I can't find a breakthrough to simplify it.
Thanks for your help.
 

Attachments

  • #6
610
398
[itex]A\ln {f(x)} = \ln {f(x)}^A[/itex] is something to keep in mind. I am using ln notation since I'm not used to logs other than base e, but regardless, the property holds. I noticed that you might not be aware of the property.
Furthermore - when [itex]X^{f(x)} = X^{g(x)} [/itex]?
What can you do to convert [itex]\sqrt 3[/itex] to a power of X?

Another thing to point out - when you deal with logs, you should also define your domain in which you can search for solutions. The initial problem can only have solutions when x + 2 > 0 and you also have a square root of x9 - which only exists in real numbers if x is non-negative.

The reason why this is important is if you do find a solution to the problem and you do not check if the solution fits into the domain, then you haven't solved the problem. A solution might not even exist, you always have to keep that possibility in mind.
 
Last edited:
  • #7
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728

Homework Statement


The equation is as follows:
23w3z0l.gif


Homework Equations


What technique shall I use to solve the equation?

The Attempt at a Solution


Tried to solve it in any way that went through my mind, including making a substitution of terms but couldn't quite get the solution or find a path towards it.
Often we solve such problems --- meaning problems whose solutions are not at all easy to see or get --- in two stages:

(1) Plot both sides on graph paper to get a rough idea of where the two curves ##y = x^{3 \log_3 (x+2)}## and ##y = \sqrt{3 x^9}## cross (if ever).
(2) Using the rough location obtained in (1), implement a numerical equation-solving routine to home in on a more accurate answer.

In this case, step (1) leads to a somewhat tricky situation: it seems to produce a solution that is almost too easy and too good to be true. In fact, it IS too good to be true; you need to actually do step (2) to get a genuine solution.

What about step (2)? You can use numerous methods manually, or you can nowadays use readily available computer tools. For example, you can use the EXCEL Solver tool if you have EXCEL on your computer. Or, you can use Maple or Mathematica if you have access to them (or other tools such as Maxima, etc.). If none of these are available, you can use the free on-line tool Wolfram Alpha (which is like Mathematica Lite) to deal with your problem. Alternatively, you can use one of the many free on-line equation-solving packages---just Google "solving equations" or something similar.
 
  • Like
Likes BruceW and berkeman
  • #9
22
0
What can you do to convert [itex]\sqrt 3[/itex] to a power of X?
What? If I had known the answer I wouldn't have asked it here. I do keep in mind all the rest of what you wrote, but that [itex]\sqrt 3[/itex] can simplify everything if I can represent it as a power of x in some form... but what form?

I have tried x=0 and it does solve it, so it is one certain solution, but there are two other solutions, which shouldn't look pretty (actually one is x=4, which is) but what matters most is how can I reach them.

Ray, you get it too complicated in this particular circumstances I'm afraid.. it should be too good to be true when solving it using algebraic technique as well.
 
  • #10
12,527
6,318
I was thinking that using log base 3 would help but it still doesn't fall apart cleanly when you use that technique. You'd take the log base 3 of both sides which reduces the sqrt of 3 to 1/2.
 
  • #11
35,269
11,543
x=4 is not a solution. It is close to one, but it is not a solution.

I fear there is no algebraic way to find the other two solutions in a reasonable form.
 
  • #12
Ray Vickson
Science Advisor
Homework Helper
Dearly Missed
10,706
1,728
What? If I had known the answer I wouldn't have asked it here. I do keep in mind all the rest of what you wrote, but that [itex]\sqrt 3[/itex] can simplify everything if I can represent it as a power of x in some form... but what form?

I have tried x=0 and it does solve it, so it is one certain solution, but there are two other solutions, which shouldn't look pretty (actually one is x=4, which is) but what matters most is how can I reach them.

Ray, you get it too complicated in this particular circumstances I'm afraid.. it should be too good to be true when solving it using algebraic technique as well.
You asked how to solve the equation, and I told you what you need to do. Drawing a graph is not complicated. Using a computerized solution package is not complicated. Before dismissing these steps you should at least give them a try.

I think you should abandon the attempt to get an algebraic solution. If the problem is exactly as you stated it, I suspect there IS NO algebraic solution, and you MUST resort to numerical methods. There really is no choice.

BTW: as has already been stated, x = 4 is not a solution (but it "looks like" one on a wide-ranging graph); that is what I meant when I said "too good to be true". If you draw a more detailed graph focussed on a region around x = 4 you will see the true situation.
 
Last edited:

Related Threads on Need some help with solving a logarithmic equation

  • Last Post
Replies
6
Views
3K
  • Last Post
Replies
6
Views
735
Replies
7
Views
1K
Replies
11
Views
2K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
5
Views
2K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
3
Views
1K
Top