Solve for x: Log x + (Log x)^2=0

[FatLouieXVI]

log x + (log x)^2 = o

It is very unclear to me how to solve this. I have managed to find x = 1, but cannot find x = 1/10. Also I have no idea if i am doing it right. How do i solve for x?

By log x i mean the common log

 

[Mark44]

This is a quadratic in log x. Further, there is no constant term in this quadratic, so it can be factored.

 

[FatLouieXVI]

log₄ x² = (log₄ x) ²

why does log₄ x = 0

i have found that the x = 16, but i am confused of why x also is 1?

 

[Bohrok]

Where did you get log₄ with base 4? If one of the answers is 1/10 as you gave in your first post, you're working with log₁₀ with base 10. Try factoring the left hand side of the equation.

 

[HallsofIvy]

log₄ x² = (log₄ x) ²

why does log₄ x = 0

i have found that the x = 16, but i am confused of why x also is 1?

log_4(x^2)= 2 log_4(x) so your equation is the same as 2log_4(x)= log_4(x). Subtracting log_4(x) from both sides, you get log_4(x)= 0 which has x= 1 as its only solution.

x= 16 is NOT a solution. 16= 4^2 so 16^2= (4^2)^2= 4^4. log_4(16^2)= 4 while log_4(16)= 2. They are NOT equal.

Did you mean (log_4(x))^2= log_4(x)? You can write that as (log_4(x))^2- log_4(x)= log_4(x)(log_4(x)- 1)= 0. Then either log_4(x)= 0, with gives x= 1, or log_4(x)- 1= 0 so that log_4(x)= 1 and x= 4. But x= 16 is still not a solution.

 

[FatLouieXVI]

sorry to clarify it is a whole new equation. The second post helped me figure the original equation out. My fault for not saying its a new equation

 

[Elucidus]

log₄ x² = (log₄ x) ²

why does log₄ x = 0

i have found that the x = 16, but i am confused of why x also is 1?

log_4(x^2)= 2 log_4(x) so your equation is the same as 2log_4(x)= log_4(x). Subtracting log_4(x) from both sides, you get log_4(x)= 0 which has x= 1 as its only solution.

x= 16 is NOT a solution. 16= 4^2 so 16^2= (4^2)^2= 4^4. log_4(16^2)= 4while log_4(16)= 2. They are NOT equal.<br /> <br /> Did you mean (log_4(x))^2= log_4(x)? You can write that as (log_4(x))^2- log_4(x)= log_4(x)(log_4(x)- 1)= 0. Then <b>either</b> log_4(x)= 0, with gives x= 1, or log_4(x)- 1= 0 so that log_4(x)= 1 and x= 4. But x= 16 is still not a solution.

<br /> <br /> x = 16 <i>is</i> a solution of \log_4(x^2) = (\log_4 x)^2. Both sides equal 4.<br /> <br /> HallsofIvy's comment though gives the hint for turning this into a (factorable) quadratic equation in log<sub>4</sub><i>x</i> which has 2 solutions, namely 1 and 16.<br /> <br /> --Elucidus

 

[HallsofIvy]

Ah, I missed the square on the x on the left side! I need to get my eyes examined!