Logaritm problem

[Loki123]

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?

 

[Deleted member 4993]

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

Image of ORIGINAL problem?

 

[BigBeachBanana]

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

Consider the domains of [imath]\log_{100}x^2[/imath] vs. [imath]2\log_{100}x[/imath]

 

[Loki123]

Image of ORIGINAL problem?

I just saw your last reply, so I didn't include it in here. I will from now on.

 

[Loki123]

Consider the domains of [imath]\log_{100}x^2[/imath] vs. [imath]2\log_{100}x[/imath]

First x can't be 0
Second x>0

 

[BigBeachBanana]

First x can't be 0
Second x>0

Correct. They are not the same function when you change the form. You omitted half of the domain when you use the second form.

 

[pka]

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

[imath]\log_{100}(x^2)=\dfrac{2\log(x)}{2\log(10}=\dfrac{\log(x)}{\log(10)}[/imath]
[imath]\log_{10}(3x+13)=\dfrac{\log(3x+13)}{\log(10)}[/imath]
So that [imath]\log_{100}(x^2)+ \log_{10}(3x+13)-1=\log\left(\dfrac{3x^2+13x}{10}\right)=0[/imath]

[imath][/imath]

 

[Loki123]

Correct. They are not the same function when you change the form. You omitted half of the domain when you use the second form.

interesting, so when I change forms, besides logarithms, I have to keep in mind the domains?

 

[pka]

interesting, so when I change forms, besides logarithms, I have to keep in mind the domains?

Well not necessarily. Use the change of base: [imath]\large{\log_B(x)=\dfrac{\log(x)}{\log(B)}}[/imath]

 

[Loki123]

Well not necessarily. Use the change of base: [imath]\large{\log_B(x)=\dfrac{\log(x)}{\log(B)}}[/imath]

while I do understand that, I am not sure I would remember to use it.

 

[BigBeachBanana]

interesting, so when I change forms, besides logarithms, I have to keep in mind the domains?

Yes, to preserve the domain, it's necessary to introduce the absolute value.
[math]\log_{100}x^2=2\log_{100}|x|[/math]