Mastering Logarithms: Simplifying a Tricky Problem in Just a Few Steps

  • Thread starter Thread starter Lancelot59
  • Start date Start date
AI Thread Summary
The discussion revolves around solving the logarithmic equation log(3-x) + log(3+x) = log(5). Participants clarify that to simplify the equation, one can use the property of logarithms that allows combining logs of the same base, leading to log(9-x^2) = log(5). This implies that 9-x^2 must equal 5, resulting in the solutions x = ±2. It's emphasized that all logarithms in an equation must have the same base to apply these properties effectively. The conversation concludes with a note on the importance of understanding logarithmic properties for solving similar problems.
Lancelot59
Messages
640
Reaction score
1
Well this is my worst area. Not sure why though. Here's the latest problem I'm stuck with.

Homework Statement


log(3-x) + log(3+x) = log(5)


Homework Equations





The Attempt at a Solution


log(9-x^2) = log(5)

The answer is plus/minus 2, but I'm really not sure on where to go from here.

This is the stuck point.
 
Physics news on Phys.org
Well, it looks like you can take 10^ of each side... see where that gets you.
 
Last edited:
There is no need to change the base of the equation. Think of it as a basic polynomial when equal to zero.
 
Last edited:
Could you give me an example? I'm not really sure what you mean by that.
 
Are you using Log to represent log base 10, or the natural log?

Assuming you are using the former, you are essentially given an equation f(x)=g(x)...If this equation holds, then so must the equation 10^{f(x)}=10^{g(x)}[/tex]...Use that with f(x)=Log[9-x^2] and g(x)=Log[5].
 
Since you're raising the same base of 10 to some number x, you can treat the equation as if there were no logarithmic notation. Using the addition property of logarithms, and the fact that no change of base is needed;

\log(x-3)+\log(x+3)=\log(5)

\log(9-x^2)=\log(5)

9-x^2=5
 
Last edited:
Sorry guys, I forgot to put in the "solve for x" part of the question. But this did enlighten me. For the longest time I thought that the only way to get a log was by using that method where there is a base and an exponent equalling another number.

My teacher didn't explain this very well.

From what I understand so far to use addition, and subtraction rules the logs need to be of the same base.

And to remove logs from certain numbers all logs in the equation need to have the same base.

Is this correct?
 
What is the definition of "logarithm"?
 
The inverse of an exponential function.
 
  • #10
Lancelot59 said:
From what I understand so far to use addition, and subtraction rules the logs need to be of the same base.

And to remove logs from certain numbers all logs in the equation need to have the same base.

Is this correct?

Hi Lancelot59! :smile:

Yes, logs in the same equation need to be of the same base (if they're not, you'll have to convert some of them).

If you're given an equation with several logs in, they will be of the same base.
Lancelot59 said:
log(3-x) + log(3+x) = log(5)

log(9-x^2) = log(5)

The answer is plus/minus 2, but I'm really not sure on where to go from here.

An equation like this, adding nothing but logs, will have the same result in any base of logs.

That's because logax = logx/loga …

so just divide that equation by log a and you get

loga(3-x) + loga(3+x) = loga(5).

In other words, your solution log(9-x^2) = log(5) works in any base, and proves that 9-x^2 = 5. :smile:
 
  • #11
So I can eliminate the logs from the equation and solve for x then. Awesome!

Thanks a lot! That really helped.
 
  • #12
You need to use the properties of logarithms.

You can find them http://people.hofstra.edu/Stefan_Waner/Realworld/calctopic1/logs.html , and also some good examples.

In this particular case the property loga(xy)= logax + logay should be used.

Regards.
 
Last edited by a moderator:
  • #13
I put it together and managed to solve it. Although I'm still stuck on all the other questions.
 
  • #14
Lancelot59 said:
… Although I'm still stuck on all the other questions.

Get some sleep :zzz: then start another thread :smile:
 
Back
Top