Logarithm question

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  • #1
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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 !
 

Answers and Replies

  • #2
87
1
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
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Well, I thought that saying log8(x)*log8(x) would be different than log8(x)^2

Ill keep waiting

Thank you
 
  • #4
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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
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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
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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 !
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
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But with the first one could I say something like : 2*(log8(x)) ??? ty
 
  • #8
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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
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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.


I don't see why not. What's in brackets is considered a single entity.
See above.
 
  • #10
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(log8(x))2 ≠ 2*log8(x) !!
Of course. :tongue:
 
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  • #11
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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
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Ok, but what is the meaning of : (log8(x))^2

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

is it (log8(x))(log8(x)) ???
Yes, this is clear.
 
  • #14
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And how would you simplify it ??

log8 x^log8(x)

Is this valid ???
 
  • #15
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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
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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
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Ok, but what is the meaning of : (log8(x))^2

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

And how would you simplify it ??

log8 x^log8(x)

Is this valid ???
Not if you mean (log8(x))2
 
  • #18
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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
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Yes.



Not if you mean (log8(x))2
Ahh..... WEll... Thank you !
 
  • #20
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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
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5,301
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
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u^2+2u+1=0

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

After Im not sure of understanding what you mean... Sorry xD
 
  • #23
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u^2+2u+1=0

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

After Im not sure of understanding what you mean... Sorry xD
Since u = log8x, you have -1 = log8x.
 
  • #24
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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
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ok thanks, i solved it yesterday !
 

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