# Is there a rule that states that I should not divide in this scenario?

• Callmelucky
In summary: You have to use ##\log_2 x^2 =2\cdot \log |x|## since the logarithm doesn't allow negative arguments. Therefore you end up with ##2^{1/2}=\sqrt{2}=|x|## which includes both signs!
Callmelucky
Homework Statement
Write the coordinates of points where ##f(x)=\log _{2} x^2 -1## intersects x and y axis
Relevant Equations
##\log _{a} x^n = n\times \log _{a} x##
So basically this is how I solved this problem:
1. ##f(x)=\log _{2} x^2 - 1##
2. ##0=\log _{2} x^2 -1 ##
3. ##1= 2\times \log _{2} x##
4. ##\frac{1}{2}= \log _{2} x##
5. ##2^{\frac{1}{2}}=x=\sqrt{2}##

So I wrote coordinates to be (##\sqrt{2}##, 0)

But apparently, that is not the only solution. There should be another answer with a negative sign so (##\pm\sqrt{2}##, 0) would be a complete solution for points at which graphs cross the x line. There are no points where the graph crosses the y-line.
This is how it's solved in the textbook(pic in attachments). And I understand that it's correct because the graph really does cross the x-line in those points.
So, my question is, is there a rule I am not aware of that states that I can't divide an equation with n(exponent of an argument moved down to the front)?

Thank you.

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Last edited:
What is the domain of your function? The problem is basically in the 3rd step.

Callmelucky
weirdoguy said:
What is the domain of your function? The problem is basically in the 3rd step.
I didn't learn about domain, in Croatia we learn that in 4th grade of high school, I am in 2nd right now. Are you saying that I can't move 2 down if it's original place is up?

You can, but it will be ##2\log_2|x|## instead of what you wrote. That's because your ##x##'s can be negative, eg: for ##x=-2## we have ##\log_2(-2)^2=\log_24=2##

Callmelucky
weirdoguy said:
You can, but it will be ##2\log_2|x|##, because your ##x##'s can be negative.
Ohh, that make sense, should I always do that when moving 2 down? Because I haven't done that once so far

In the cases of functions like this one in your example yes. The rule: ##\log a^n=n\log a## works only for positive ##a##, and in the function you gave your "##a##" (which is ##x##) could be negative.

Callmelucky
weirdoguy said:
In the cases of functions like this one in your example yes. The rule: ##\log a^n=n\log a## works only for positive ##a##, and in the function you gave your "##a##" (which is ##x##) could be negative.
Okay, thank you very much. :D

Callmelucky said:
Homework Statement: Write the coordinates of points where ##f(x)=\log _{2} x^2 -1## intersects x and y axis
Relevant Equations: ##\log _{a} x^n = n\times \log _{a} x##

So basically this is how I solved this problem:
1. ##f(x)=\log _{2} x^2 - 1##
2. ##0=\log _{2} x^2 -1 ##
3. ##1= 2\times \log _{2} x##
4. ##\frac{1}{2}= \log _{2} x##
5. ##2^{\frac{1}{2}}=x=\sqrt{2}##

So I wrote coordinates to be (##\sqrt{2}##, 0)
From line 2, you have ##\log_2(x^2) = 1 \Rightarrow x^2 = 2 \Rightarrow x= \pm \sqrt 2##
Callmelucky said:
But apparently, that is not the only solution. There should be another answer with a negative sign so (##\pm\sqrt{2}##, 0) would be a complete solution for points at which graphs cross the x line. There are no points where the graph crosses the y-line.

Callmelucky
Mark44 said:
From line 2, you have ##\log_2(x^2) = 1 \Rightarrow x^2 = 2 \Rightarrow x= \pm \sqrt 2##
yeah, but I was confused why the method I was using so far isn't working anymore. Thank you.

Callmelucky said:
yeah, but I was confused why the method I was using so far isn't working anymore. Thank you.
It's because your relevant equation -- ##\log_a (x^n) = n\log_a(x)## -- is valid only for x > 0. Although ##x = \sqrt 2## is a solution of equations 2 and 4, ##x = -\sqrt 2## is also a solution of the original equation.

Callmelucky said:
yeah, but I was confused why the method I was using so far isn't working anymore. Thank you.
You have to use ##\log_2 x^2 =2\cdot \log |x|## since the logarithm doesn't allow negative arguments. Therefore you end up with ##2^{1/2}=\sqrt{2}=|x|## which includes both signs!

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