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 $16^2= (4^2)^2= 4^4$. [itex]log_4(16^2)= 4[itex] 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.

x = 16 is a solution of $\log_4(x^2) = (\log_4 x)^2.$ Both sides equal 4.

HallsofIvy's comment though gives the hint for turning this into a (factorable) quadratic equation in log₄x which has 2 solutions, namely 1 and 16.

--Elucidus

 

[HallsofIvy]

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