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

[chemistry1]

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 !

 

[goldust]

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!

 

[chemistry1]

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

Ill keep waiting

Thank you

 

[goldust]

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/

 

[chemistry1]

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]

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.

 

[chemistry1]

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

 

[goldust]

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]

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.
log_a(x²) = 2 log_a(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.

 

[goldust]

(log₈(x))² ≠ 2*log₈(x) !

Of course.

 

[economicsnerd]

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.

 

[chemistry1]

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

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

 

[goldust]

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

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

Yes, this is clear.

 

[chemistry1]

And how would you simplify it ??

log8 x^log8(x)

Is this valid ?

 

[goldust]

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.

 

[chemistry1]

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]

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]

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.

 

[chemistry1]

Yes.



Not if you mean (log₈(x))²

Ahh... WEll... Thank you !

 

[chemistry1]

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]

Just replace log_8](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.

 

[chemistry1]

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

 

[goldust]

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.

 

[Mark44]

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.

 

[chemistry1]

ok thanks, i solved it yesterday !