Bloody Logs (1 Viewer)

[spatula232]

Was doing a past paper and had the Q: ln(e^(x+y))² -- They had the answer 2(x+y); but I thought you could only bring the power down if it was within the bracket. Wouldn't it just be (x+y).ln(e)^2 = x+y???

An explanation would be great

 

[VBN2470]

Answer should be , but if the 2 was inside the bracket as such then the answer would be

 

[spatula232]

So then why can you pull the whole (x+y)^2 down? This is the part that's confusing me because I thought you could only do that if it was in the bracket - and then that would be the case for what you said for the second time

What I'm asking in essence is when does the power log rule apply? Because I thought it was only for when the power was within the bracket, like ln(e^2(x+y))

 

[VBN2470]

I'm assuming the notation is as such:

 

[VBN2470]

So what I mean is that they actually mean

 

[VBN2470]

The power log rule applies. I am not actually pulling down, I am pulling down and squaring it.

 

[rand_althor]

What I'm asking in essence is when does the power log rule apply? Because I thought it was only for when the power was within the bracket, like ln(e^2(x+y))

The rule is

You are right - it is only when the power is in the bracket, i.e.

 

[spatula232]

Thanks

 

[VBN2470]

E.g. If we want to simplify , this actually means that we want to find

 

[InteGrand]

Was doing a past paper and had the Q: ln(e^(x+y))² -- They had the answer 2(x+y); but I thought you could only bring the power down if it was within the bracket. Wouldn't it just be (x+y).ln(e)^2 = x+y???

An explanation would be great