Logarithms: (log8(x))^2 vs log8(x)^2

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In summary, the conversation discusses the difference in notation between (log8(x))^2 and log8(x)^2. The consensus is that they mean the same thing, with the latter being ambiguous. The conversation also touches on the simplification of (log8(x))^2 and solving an equation involving logarithms.
  • #1
chemistry1
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Hi, quick question.

Is there a difference in notation when I say :

(log8(x))^2

and

log8(x)^2

Is it different or the same ? Thank you !
 
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  • #2
I would say they mean the same thing. log8(x^2) would be different. To me, at least, log8(x)^2 means squaring log8(x). Hope that helps!
 
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  • #3
Well, I thought that saying log8(x)*log8(x) would be different than log8(x)^2

Ill keep waiting

Thank you
 
  • #4
You can try it out on a calculator and see if there is any difference between (log8(x))^2 and log8(x)^2. :biggrin:

http://web2.0calc.com/
 
  • #5
Yeah, it seems to say that it's the samething. COuld you try it on this one, I have doubts :https://www.mathway.com/

Thank you
 
  • #6
chemistry1 said:
Hi, quick question.

Is there a difference in notation when I say :

(log8(x))^2

and

log8(x)^2

Is it different or the same ? Thank you !

The first one is clear, but the second one is ambiguous. It could be interpreted as either (log8(x))2, or as log8(x2).

Parentheses should be used to make your meaning clear.
 
  • #7
But with the first one could I say something like : 2*(log8(x)) ? ty
 
  • #8
chemistry1 said:
But with the first one could I say something like : 2*(log8(x)) ? ty

I don't see why not. What's in brackets is considered a single entity.
 
  • #9
chemistry1 said:
But with the first one could I say something like : 2*(log8(x)) ? ty
(log8(x))2 ≠ 2*log8(x) ! That's not how the log properties work.
loga(x2) = 2 loga(x). This isn't what you have in the first example.


goldust said:
I don't see why not. What's in brackets is considered a single entity.
See above.
 
  • #10
Mark44 said:
(log8(x))2 ≠ 2*log8(x) !

Of course. :tongue:
 
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  • #11
As Mark44 said, just use parentheses to make things unambiguous.
- [itex](\log_8 x)^2[/itex] means one thing.
- [itex]\log_8 (x^2)[/itex] means another thing (which happens to be the same thing that [itex]2\log_8 x[/itex] means).
- Nobody is stopping you from writing [itex]\log_8(x)^2[/itex], but I wouldn't write it, because it's not crystal clear what it means.
 
  • #12
Ok, but what is the meaning of : (log8(x))^2

is it (log8(x))(log8(x)) ?
 
  • #13
chemistry1 said:
Ok, but what is the meaning of : (log8(x))^2

is it (log8(x))(log8(x)) ?

Yes, this is clear.
 
  • #14
And how would you simplify it ??

log8 x^log8(x)

Is this valid ?
 
  • #15
chemistry1 said:
And how would you simplify it ??

log8 x^log8(x)

Is this valid ?

To me, this is the same as log8 (x^log8(x)), because you can only have a single input with the function log8.
 
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  • #16
Look, here's the reason why I'm asking the question :

(log8(x))^2+2(log(x))+1=0

I'm not able of finding a way to solve this. The closest answer I got was 1/64. Can somebody help me ?
 
  • #17
chemistry1 said:
Ok, but what is the meaning of : (log8(x))^2

is it (log8(x))(log8(x)) ?
Yes.

chemistry1 said:
And how would you simplify it ??

log8 x^log8(x)

Is this valid ?

Not if you mean (log8(x))2
 
  • #18
chemistry1 said:
Look, here's the reason why I'm asking the question :

(log8(x))^2+2(log(x))+1=0

I'm not able of finding a way to solve this. The closest answer I got was 1/64. Can somebody help me ?
Did you omit the base in the second term? Should it be log8(x)?

If so, your equation is quadratic in form. If you let u = log8(x), then the equation can be written as x2 + 2x + 1 = 0, which can be factored.
 
  • #19
Mark44 said:
Yes.



Not if you mean (log8(x))2
Ahh... WEll... Thank you !
 
  • #20
Mark44 said:
Did you omit the base in the second term? Should it be log8(x)?

If so, your equation is quadratic in form. If you let u = log8(x), then the equation can be written as x2 + 2x + 1 = 0, which can be factored.
AH yes, I forgot it ! What do you mean by u= ... ?
 
  • #21
Just replace log8](x) by u in your equation. The you have a true quadratic, not one that is just quadratic in form.

Solve the equation u2 + 2u + 1 = 0 for u, and then solve for x by undoing the substitution.
 
  • #22
u^2+2u+1=0

(u+1)(u+1)=0 SOlution : u=-1

After I am not sure of understanding what you mean... Sorry xD
 
  • #23
chemistry1 said:
u^2+2u+1=0

(u+1)(u+1)=0 SOlution : u=-1

After I am not sure of understanding what you mean... Sorry xD

Since u = log8x, you have -1 = log8x.
 
  • #24
chemistry1, keep in mind that a logarithm is an exponent on a particular base. For your equation, log8(x) is -1. This means that -1 is the exponent on the base (8) that produces x.
 
  • #25
ok thanks, i solved it yesterday !
 

1. What is the difference between (log8(x))^2 and log8(x)^2?

The difference between (log8(x))^2 and log8(x)^2 is the placement of the parentheses. In (log8(x))^2, the logarithm is squared, while in log8(x)^2, the result of the logarithm is squared. This can lead to different values for the two expressions.

2. Which expression, (log8(x))^2 or log8(x)^2, should I use to solve a logarithmic equation?

Both expressions can be used to solve a logarithmic equation, but it is important to be consistent with the use of parentheses in the equation. If the equation already has parentheses around the logarithm, then use log8(x)^2. If there are no parentheses, then use (log8(x))^2.

3. How do I simplify the expression (log8(x))^2?

To simplify (log8(x))^2, you can use the logarithmic property loga(x^b) = b*loga(x). This means that (log8(x))^2 can be rewritten as 2*log8(x). Then, using the change of base formula, loga(x) = logb(x)/logb(a), (log8(x))^2 can be simplified to 2*log(x)/log(8).

4. Can (log8(x))^2 be written as log8(x^2)?

Yes, (log8(x))^2 can be written as log8(x^2). This is because of the logarithmic property loga(x^b) = b*loga(x). In this case, b = 2, so (log8(x))^2 is equivalent to 2*log8(x). This can then be rewritten as log8(x^2).

5. Is there a difference between (log8(x))^2 and log8(x^2)?

Yes, there is a difference between (log8(x))^2 and log8(x^2). In (log8(x))^2, the logarithm is squared, while in log8(x^2), the argument of the logarithm is squared. This can lead to different values for the two expressions, unless x is equal to 1 or 8.

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