Solve this Logarithmic Equation: Log2x^log2x=4

  • Thread starter jaypee
  • Start date
In summary, there is some ambiguity in the original problem. If the equation is interpreted as (log2 x)log2x = 4, then the solution x=4 is valid. However, if it is interpreted as log2(x^(log2(x)))=4, then x=4 is still a solution but it is unknown if there are any other solutions. The convention is to give priority to the exponent over the function, but it is always best to use parentheses to avoid ambiguity.
  • #1
jaypee
Can Someone please solve this for me
Log2x^log2x=4
Where 2 is the base of log and exponent.
 
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  • #2
log2 xlog 2x = 4
(log2 x)(log2 x) = 4
(log2 x)2 = 4
log2 x = ±√4 = ±2
x = 2±2 = 4,1/4
 
  • #3
lethe, I think your 1st step is invalid. I think the only answer is x=4.
 
  • #4
No, I think his first step is valid. Remember that ln(xa) = aln(x)!
 
  • #5
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
(log2 x)log2 x = 4
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)-2=1/4, not 4.

but if the original poster meant the equation that i wrote above,
log2 xlog 2x = 4
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:
log2 (1/4)-2 = log2 16 = 4

bangarang.

climbhi-
thanks for gettin my back.
 
  • #6
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?
 
Last edited:
  • #7
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 x2, what would you assume? usually, with cosine, the power takes priority. if you want the cosine function to take priority, you must write either (cos x)2 or cos2 x. so you see that normally, the exponent has higher priority in the order of operations, and i think my assumption was reasonable. i think the same argument applies to logarithms.

if you see log x3, would you really assume that it means (log x)3? no, that always means power first. if you want to raise the function to the power, it should appear in parentheses explicitly.
 
  • #8
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 ?
 
Last edited:
  • #9
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.

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
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.
 

What is a logarithmic equation?

A logarithmic equation is an equation in which the variable appears in the exponent of a logarithm.

How do you solve logarithmic equations?

To solve a logarithmic equation, you need to use the properties of logarithms to isolate the variable on one side of the equation and then solve for it using exponentiation.

What are the properties of logarithms?

The properties of logarithms include the product rule, quotient rule, power rule, and change of base rule.

What does "log" mean in mathematics?

"Log" is short for logarithm, which is a mathematical function that represents the power to which a base number must be raised to produce a given number.

What is the solution to Log2x^log2x=4?

The solution to Log2x^log2x=4 is x=16. This can be found by using the properties of logarithms to rewrite the equation as x^log2x=2^4 and then solving for x by taking the logarithm of both sides and simplifying the resulting exponential equation.

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