My problem is with the first logaritm. I would write it using the first way, however I get all answers with second way. Why doesn't the first way work and the second does? View attachment 31544
My problem is with the first logaritm. I would write it using the first way, however I get all answers with second way. Why doesn't the first way work and the second does? View attachment 31544
I just saw your last reply, so I didn't include it in here. I will from now on.Image of ORIGINAL problem?
First x can't be 0Consider the domains of [imath]\log_{100}x^2[/imath] vs. [imath]2\log_{100}x[/imath]
Correct. They are not the same function when you change the form. You omitted half of the domain when you use the second form.First x can't be 0
Second x>0
[imath]\log_{100}(x^2)=\dfrac{2\log(x)}{2\log(10}=\dfrac{\log(x)}{\log(10)}[/imath]My problem is with the first logaritm. I would write it using the first way, however I get all answers with second way. Why doesn't the first way work and the second does? View attachment 31544
interesting, so when I change forms, besides logarithms, I have to keep in mind the domains?Correct. They are not the same function when you change the form. You omitted half of the domain when you use the second form.
Well not necessarily. Use the change of base: [imath]\large{\log_B(x)=\dfrac{\log(x)}{\log(B)}}[/imath]interesting, so when I change forms, besides logarithms, I have to keep in mind the domains?
while I do understand that, I am not sure I would remember to use it.Well not necessarily. Use the change of base: [imath]\large{\log_B(x)=\dfrac{\log(x)}{\log(B)}}[/imath]
Yes, to preserve the domain, it's necessary to introduce the absolute value.interesting, so when I change forms, besides logarithms, I have to keep in mind the domains?