# Logarithm question

## Main Question or Discussion Point

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 !

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

Ill keep waiting

Thank you

You can try it out on a calculator and see if there is any difference between (log8(x))^2 and log8(x)^2. http://web2.0calc.com/

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

Mark44
Mentor
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.

But with the first one could I say something like : 2*(log8(x)) ??? ty

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.

Mark44
Mentor
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.

(log8(x))2 ≠ 2*log8(x) !!
Of course. :tongue:

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As Mark44 said, just use parentheses to make things unambiguous.
- $(\log_8 x)^2$ means one thing.
- $\log_8 (x^2)$ means another thing (which happens to be the same thing that $2\log_8 x$ means).
- Nobody is stopping you from writing $\log_8(x)^2$, but I wouldn't write it, because it's not crystal clear what it means.

Ok, but what is the meaning of : (log8(x))^2

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

Ok, but what is the meaning of : (log8(x))^2

is it (log8(x))(log8(x)) ???
Yes, this is clear.

And how would you simplify it ??

log8 x^log8(x)

Is this valid ???

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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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 ???

Mark44
Mentor
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

Mark44
Mentor
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.

Yes.

Not if you mean (log8(x))2
Ahh..... WEll... Thank you !

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= ... ???

Mark44
Mentor
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.

u^2+2u+1=0

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

After Im not sure of understanding what you mean... Sorry xD

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.

Mark44
Mentor
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.

ok thanks, i solved it yesterday !