can someone please me where I went wrong on this question, in the solutionbank they sub in the conditions half way through the working out, i'm not sure why my method isn't valid?
can someone please me where I went wrong on this question, in the solutionbank they sub in the conditions half way through the working out, i'm not sure why my method isn't valid?
the answer is meant to be (8+4y)/(2-y)
when you removed the logs on the first page near the end you added the terms when really the constant is multiplied by the other term
can someone please me where I went wrong on this question, in the solutionbank they sub in the conditions half way through the working out, i'm not sure why my method isn't valid?
the answer is meant to be (8+4y)/(2-y)
In the line where you went from =ln∣secx∣+c to =ln∣secx∣+lnA you then left it as a +A when instead you should have done ln∣secx∣+lnA=ln∣Asecx∣
This makes your solution 2−y2+y=Asecx=secx+A. You capiche?
In the line where you went from =ln∣secx∣+c to =ln∣secx∣+lnA you then left it as a +A when instead you should have done ln∣secx∣+lnA=ln∣Asecx∣
This makes your solution 2−y2+y=Asecx=secx+A. You capiche?
thank youuuuuuuuu i see where I went wrong now
but out of interest, why have I got to combine the In(secx) with InA before I can remove the logs? have all the log terms got to be grouped up before you can then move the log part?
but out of interest, why have I got to multiply the In(secx) with InA before I can remove the logs? have all the log terms got to be grouped up before you can then move the log part?
Well basically what you're doing when you remove the logs is applying e to the power of each side. You technically don't have to explicitly write the log terms grouped up first, but you have to know what will happen to them when you make this manipulation.
but out of interest, why have I got to multiply the In(secx) with InA before I can remove the logs? have all the log terms got to be grouped up before you can then move the log part?
Yes, in general when you have something like lnA=lnB+c then you're taught to group everything up as lnA+lnBec⇒A=Bec. The real reason why you're doing this is because you're taking the exp (exponential) of both sides of the equation.
So it's a bit like when you have x3=a+b⇒x=(a+b)1/3=a1/3+b1/3, here you take the exponential of both sides to get:
lnA=lnB+c⇒elnA=elnB+c=elnBec⇒A=Bec using the fact that elnx=x.
Well basically what you're doing when you remove the logs is applying e to the power of each side. You technically don't have to explicitly write the log terms grouped up first, but you have to know what will happen to them when you make this manipulation.
Well basically what you're doing when you remove the logs is applying e to the power of each side. You technically don't have to explicitly write the log terms grouped up first, but you have to know what will happen to them when you make this manipulation.
Yes, in general when you have something like lnA=lnB+c then you're taught to group everything up as lnA+lnBec⇒A=Bec. The real reason why you're doing this is because you're taking the exp (exponential) of both sides of the equation.
So it's a bit like when you have x3=a+b⇒x=(a+b)1/3=a1/3+b1/3, here you take the exponential of both sides to get:
lnA=lnB+c⇒elnA=elnB+c=elnBec⇒A=Bec using the fact that elnx=x.
thank you so much both of you!! makes sense to me now
zacken in there with a very detailed answer, you should be teaching maths you are really good at explaining things