- #1

jaypee

Can Someone please solve this for me

Log2x^log2x=4

Where 2 is the base of log and exponent.

Log2x^log2x=4

Where 2 is the base of log and exponent.

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- Thread starter jaypee
- Start date

- #1

jaypee

Can Someone please solve this for me

Log2x^log2x=4

Where 2 is the base of log and exponent.

Log2x^log2x=4

Where 2 is the base of log and exponent.

- #2

- 653

- 0

(log

(log

log

x = 2

- #3

- 508

- 0

lethe, I think your 1st step is invalid. I think the only answer is x=4.

- #4

climbhi

No, I think his first step is valid. Remember that ln(x^{a}) = aln(x)!

- #5

- 653

- 0

Originally posted by arcnets

lethe, I think your 1st step is invalid. I think the only answer is x=4.

arcnets-

there is some ambiguity in the original post. if she meant

(log

then my solution is incorrect. in this case, the equation is transcendental, and i can t solve it. one can easily verify that x=4 is still a solution, but i have no idea if there are other solutions. i guess i could graph it and look for more. x=1/4 is not a solution in that case, because (-2)

but if the original poster meant the equation that i wrote above,

log

then my solution is correct, and i have found all solutions. i hope that this is what the poster intended, because this is a soluble equation. if you don t believe that x=1/4 is a valid solution, all you have to do is check:

log

bangarang.

climbhi-

thanks for gettin my back.

- #6

- 508

- 0

Oops!

Obviously I misinterpreted the problem. Seeing no brackets, I thought that the problem was (log2(x))^(log2(x))=4.

While lethe's solution is correct for log2(x^(log2(x)))=4.

IOW, I thought that a functional symbol (like 'log') has priority over a power. I must have been wrong.

Is there such a convention? Any comments?

Obviously I misinterpreted the problem. Seeing no brackets, I thought that the problem was (log2(x))^(log2(x))=4.

While lethe's solution is correct for log2(x^(log2(x)))=4.

IOW, I thought that a functional symbol (like 'log') has priority over a power. I must have been wrong.

Is there such a convention? Any comments?

Last edited:

- #7

- 653

- 0

Originally posted by arcnets

Oops!

Obviously I misinterpreted the problem. Seeing no brackets, I thought that the problem was (log2(x))^(log2(x))=4.

While lethe's solution is correct for log2(x^(log2(x)))=4.

IOW, I thought that a functional symbol (like 'log') has priority over a power. I must have been wrong.

Is there such a convention? Any comments?

well, if you saw this: cos x

if you see log x

- #8

- 508

- 0

lethe, what you say makes sense to me.

Which leaves the question, where exactly do we have to place functional symbols in the order of priority?

Look e.g. at sin ab ^ a sin b.

OK, power first, but it could still mean

sin (ab^a) * sin b

or

sin(ab^a * sin b).

Any comments? Function first OR multiplication first ?

Which leaves the question, where exactly do we have to place functional symbols in the order of priority?

Look e.g. at sin ab ^ a sin b.

OK, power first, but it could still mean

sin (ab^a) * sin b

or

sin(ab^a * sin b).

Any comments? Function first OR multiplication first ?

Last edited:

- #9

- 653

- 0

thus cos ωt is cos(ωt)

however, my intuition about your example would lead me to choose the first choice, even though that is in violation of the rules of order of operation. very rarely is it natural to have the sine of a sine, so i would not naturally assume that is what is meant.

in such a case, i would certainly use parens.

- #10

climbhi

Originally posted by lethe

i usually write multiplications to the left of the function, to avoid ambiguity. anything multiplied on the right goes in the functions argument. so multiplication before function.

This seems to be the pretty standard way of doing things and makes the most sense. I remember when I was first taking calculus and we were learning the product rule and other rules for derivatives. My teacher would always differentiate say the sine function and then multiply the derivative of what was on the inside on the right of the function instead of the left. It was very confusing for me becuase I always kept trying to make it part of the function.

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