# Need help making X the subject

• Nov 1st 2012, 07:52 AM
funbuoj
Need help making X the subject
Hi all, the attached document shows 7 questions to "rearrange to make x the subject". I have answers for some of them but other I'm completely stumped on where to even begin. Here are my answers so far

1)i) y= mx+c
y-c= mx
y-c/m = x

ii) y=mx+cx+c
y-c = mx+cx
I'm not sure where to go after this, I would come out with a final answer of y-c/mc = 2x?

iii) y=x+c/a/b - once again i'm not even sure where to go with this to start with

iv) y=x^2+c
y-c=x^2
sqroot(y-c)=x

v) y=e^Lx -1
y+1 = e^Lx
not sure where to go with this, I imagine you use logs somehow

vi) y=xcos^2 + xsin^2
Completely stumped

vii) y=log(e^x)
Not sure where to begin

All help is apprecaited! many thanks
• Nov 1st 2012, 10:04 AM
HallsofIvy
Re: Need help making X the subject
Quote:

Originally Posted by funbuoj
Hi all, the attached document shows 7 questions to "rearrange to make x the subject". I have answers for some of them but other I'm completely stumped on where to even begin. Here are my answers so far

1)i) y= mx+c
y-c= mx
y-c/m = x

Be careful how you write this! You are dividing y- c by m so that would be written (y- c)/m= x. What you wrote would normally be interpreted as y- (c/m)= x which is not correct.

Quote:

ii) y=mx+cx+c
y-c = mx+cx
I'm not sure where to go after this, I would come out with a final answer of y-c/mc = 2x?
No, mx+ cx= (m+c)x. To solve (m+ c)x= y- c, you get x= (y- c)/(m+ c)

Quote:

iii) y=x+c/a/b - once again i'm not even sure where to go with this to start with
I don't know what you mean by this. Is that supposed go be y= ((x+ c)/a)/b or y= x+ (c/a)/b?

Quote:

iv) y=x^2+c
y-c=x^2
sqroot(y-c)=x
The equation x^2= y- c has two solutions, x= sqrt(y-c), which you give and x= -sqrt(y- c).

[quote]v) y=e^Lx -1
y+1 = e^Lx
not sure where to go with this, I imagine you use logs somehow
Quote:

Assuming you mean e^(Lx), yes. The definition of the (natural) logarithm is that it is the inverse function to e^x: log(e^x)= x. So if you take the logarithm of both sides, log(y+ 1)= log(e^(Lx))= Lx. Now divide both sides by L.

Quote:

vi) y=xcos^2 + xsin^2
Completely stumped
So am I! I have no idea what "cos" and "sin", could possibly mean without an argument! Is it possible you meant cos^2(x)+ sin^2(x) or xsin^2(x)+ xcos^2(x)? If it is either of those, use the fact that sin^2(x)+ cos^2(x)= 1.

vii) y=log(e^x)
Not sure where to begin

All help is apprecaited! many thanks