another logarithm question.

[Doubell]

1. The problem statement, all variables and given/known data

by substituting y = log2x solve for x in the following equation:

√log2x = logs2√x
2. Relevant equations

logab=c then a^c = b
3. The attempt at a solution

if y = log2x then the equation becomes √y = log2 x^1/2
this implies √y = 1/2 log2x which simplifies to √y = 1/2 y
[√y]^2 = [ 1/2 y]^2
y = (y^2)/4
4y = y^2
4y-y^2 = 0
y(4-y) = 0
4-y = 0
y = 4
if y = 4 and y = log2x then 4 = log2x
if loga b = c then a ^c = b
this implies that 2^4 = x and x = 16. anyone agrees with this solution

[SammyS]

1. The problem statement, all variables and given/known data

by substituting y = log2x solve for x in the following equation:

√log2x = logs2√x
2. Relevant equations

logab=c then a^c = b
3. The attempt at a solution

if y = log2x then the equation becomes √y = log2 x^1/2
this implies √y = 1/2 log2x which simplifies to √y = 1/2 y
[√y]^2 = [ 1/2 y]^2
y = (y^2)/4
4y = y^2
4y-y^2 = 0
y(4-y) = 0
4-y = 0
y = 4
if y = 4 and y = log2x then 4 = log2x
if loga b = c then a ^c = b
this implies that 2^4 = x and x = 16. anyone agrees with this solution

Is this the equation you're supposed to be solving \sqrt{\log_2\,x\ }=\log_2\,\sqrt{x}\ \ ?

The equation 2u2 = u , has two solutions. So does the equation 2y=\sqrt{y}\,.

Write 2u2 = u as 2u2 - u = 0, then factor out the common factor.

[HallsofIvy]

SammyS, isn't that exactly what he said he did?

Doublell, it's easy to check your answer. If x= 16 then \sqrt{x}= 4 and log_2(\sqrt{x})= log_2(4)= log_2(2^2)= 2. Of course, log_2(16)= log_2(2^4)= 4 so \sqrt{log_2(x})= \sqrt{4}= 2 also.

[SammyS]

SammyS, isn't that exactly what he said he did?

Doublell, it's easy to check your answer. If x= 16 then \sqrt{x}= 4 and log_2(\sqrt{x})= log_2(4)= log_2(2^2)= 2. Of course, log_2(16)= log_2(2^4)= 4 so \sqrt{log_2(x})= \sqrt{4}= 2 also.
Well, I admit that I didn't read his post as carefully as I should have. (I may have spent too much time working with another PH user, and some of his behaviors were contagious.) However, what I should have pointed out, is that if y(4 - y) = 0, there are two solutions for y. OP did drop the y = 0 solution.

If log2(x) = 0, then x = 1.

[Doubell]

well, i admit that i didn't read his post as carefully as i should have. (i may have spent too much time working with another ph user, and some of his behaviors were contagious.) however, what i should have pointed out, is that if y(4 - y) = 0, there are two solutions for y. Op did drop the y = 0 solution.

If log2(x) = 0, then x = 1.

i noticed that my post are not as clear as u guys eg i write log2x when in ur posts its clear to understand any advice on how i can post my questions in a similar fashion as yours?

[Mark44]

i noticed that my post are not as clear as u guys eg i write log2x when in ur posts its clear to understand any advice on how i can post my questions in a similar fashion as yours?

Please - no textspeak (e.g., u and ur). Using textspeak is a violation of forum rules.

You can write exponents and subscripts using the expanded menu that is available when you click Go Advanced. For subscripts, as in log2(x), click the X2 button and enter the subscript. (It doesn't have to be 2.)

For exponents, as in w4, click the X2 button and enter the exponent. There are a bunch of other symbols that you can use, shown to the right of the text-entry window, such as √, ≤, Ʃ, ±, and Greek letters.

[Doubell]

please - no textspeak (e.g., u and ur). Using textspeak is a violation of forum rules.

You can write exponents and subscripts using the expanded menu that is available when you click go advanced. For subscripts, as in log2(x), click the x2 button and enter the subscript. (it doesn't have to be 2.)

for exponents, as in w4, click the x2 button and enter the exponent. There are a bunch of other symbols that you can use, shown to the right of the text-entry window, such as √, ≤, Ʃ, ±, and greek letters.

thanks and i will remember no text speaking