Logarithm Question: Solving for x with Substitution

[Doubell]

Homework Statement

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

√log2x = logs2√x

Homework Equations

logab=c then a^c = b

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]

Homework Statement

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

√log2x = logs2√x

Homework Equations

logab=c then a^c = b

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]

Writing my posts more clearly

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