Year 13 Maths Help Thread

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Your first half of the LHS is right but you have brought e^...+ lny down to give lny, which isn't quite right :smile:

Remember that

x^{a + b} = x^a \times x^b

What do you mean i havent brought down lny??

Reply 881

Kevin De Bruyne

Original post by kiiten

What do you mean i havent brought down lny??

Okay, how did you get from the line with e^ln(...) to the line below? how did you get the +lny in the equation below it? :smile:

Reply 882

kiiten

Original post by SeanFM

Okay, how did you get from the line with e^ln(...) to the line below? how did you get the +lny in the equation below it? :smile:

I moved everything after e^ln in front of e^ln (e^5 - e^2 etc.). Then e^ln cancels out which leaves e^5 - e^2 etc.
lny is in the equation because that was also part of the index notation?

Reply 883

RDKGames

Study Forum Helper

Original post by kiiten

Different ques

???? I got y = 2 / ln

Where did i go wrong?

Posted from TSR Mobile

I'm not quite sure what you are trying to do but there is a whole lot simpler way to solve it if you get rid off the natural logarithm.

Just use the fact that

\ln(x)+\ln(y)=\ln(xy)=7

xy=e^7

Then rearrange for, say,

x

and we get

x=\frac{e^7}{y}

Subbing it into the first equation we get:

\frac{e^7}{y}-y=e^5-e^2

Which leads to

e^7-y^2=y(e^5-e^2)

and just solve for

y

and then

x

(edited 7 years ago)

Reply 884

Kevin De Bruyne

Original post by kiiten

I moved everything after e^ln in front of e^ln (e^5 - e^2 etc.). Then e^ln cancels out which leaves e^5 - e^2 etc.
lny is in the equation because that was also part of the index notation?

the e^5 - e^2 - y can be taken down but notice that it adds lny...

What you've done isn't completely correct is what I'm trying to say.

e^{a+b} = e^a \times e^b

where a and b are whatever you want.

Reply 885

kiiten

Original post by RDKGames

I'm not quite sure what you are trying to do but there is a whole lot simpler way to solve it if you get rid off the natural logarithm.

Just use the fact that

\ln(x)+\ln(y)=\ln(xy)=7

xy=e^7

Then rearrange for, say,

x

and we get

x=\frac{e^7}{y}

Subbing it into the first equation we get:

\frac{e^7}{y}-y=e^5-e^2

Which leads to

e^7-y^2=y(e^5-e^2)

and just solve for

y

and then

x

Thanks

- i understand what youre saying here but im not too sure how you would solve to find y.
EDIT: sorry its in pencil :3

(edited 7 years ago)

Reply 886

kiiten

Original post by SeanFM

the e^5 - e^2 - y can be taken down but notice that it adds lny...

What you've done isn't completely correct is what I'm trying to say.

e^{a+b} = e^a \times e^b

where a and b are whatever you want.

So what happens to the lny? Do you mean it doesnt work because lny isnt inside the bracket?

Reply 887

Kevin De Bruyne

Original post by kiiten

So what happens to the lny? Do you mean it doesnt work because lny isnt inside the bracket?

That is what I am trying to show you :h:

(what happens to the lny)

Like I said from the previous post,

In this case a is ln(e^5 - e^2 + y) and b is ln(y).

So e^a+b is e^(ln(e^5 - e^2 + y) * e^(lny) = ....

Now you should be able to see the difference between what you did in post 882 and what has just been done.

Reply 888

kiiten

Original post by SeanFM

That is what I am trying to show you :h:

(what happens to the lny)

Like I said from the previous post,

Ahh i see. Thank youu :biggrin:

- sorry i just wasnt getting it at all but i understand now. Just to check does this lead to:

y(e^5 - e^2 + y) = e^7
y = e^7 / (e^5 - e^2 + y)
y = e^2 - e^5 + y/(e^5 - e^2 + y)

??

Reply 889

Kevin De Bruyne

Original post by kiiten

Ahh i see. Thank youu :biggrin:

- sorry i just wasnt getting it at all but i understand now. Just to check does this lead to:

y(e^5 - e^2 + y) = e^7
y = e^7 / (e^5 - e^2 + y)
y = e^2 - e^5 + y/(e^5 - e^2 + y)

??

After your first line of the quoted post it is easier to turn it into a quadratic, as you can see you've reached a dead end.. I think.

Reply 890

kiiten

Original post by SeanFM

After your first line of the quoted post it is easier to turn it into a quadratic, as you can see you've reached a dead end.. I think.

Yep

. Ohh So:

ye^5 - ye^2 + y^2 = e^7
y^2 - ye^2 + ye^5 - e^7 = 0
y(y - e^2) + e^5 (y - e^2) = 0
(y+ e^5)(y - e^2) = 0
y = e^2

Finally, yay :smile:

- is this right? - question only asks for +ve solutions so the other bracket is invalid.

Reply 891

Kevin De Bruyne

Original post by kiiten

Yep

. Ohh So:

ye^5 - ye^2 + y^2 = e^7
y^2 - ye^2 + ye^5 - e^7 = 0
y(y - e^2) + e^5 (y - e^2) = 0
(y+ e^5)(y - e^2) = 0
y = e^2

Finally, yay :smile:

- is this right? - question only asks for +ve solutions so the other bracket is invalid.

You can always check the solutions of an equation by putting them back into the equation :smile:

Reply 892

kiiten

Original post by SeanFM

You can always check the solutions of an equation by putting them back into the equation :smile:

Yea I think i figured it out at last. The solutions work. Thanks so much for your help :biggrin:

One more question - dy/dx of (x+1)^3(x+2)^6 do you use the chain rule?
I ended up with 3(x+1)^2 6(x+2)^5 but the question wants it in the form (x+1)^2 (x+2)^5 (ax+b) ?? :s-smilie:

Reply 893

Kevin De Bruyne

Original post by kiiten

Yea I think i figured it out at last. The solutions work. Thanks so much for your help :biggrin:

One more question - dy/dx of (x+1)^3(x+2)^6 do you use the chain rule?
I ended up with 3(x+1)^2 6(x+2)^5 but the question wants it in the form (x+1)^2 (x+2)^5 (ax+b) ?? :s-smilie:

Think about the product rule too :smile:

Reply 894

kiiten

Original post by SeanFM

Think about the product rule too :smile:

Ah but im not sure about using the product rule with brackets. I tried it with (x+1)^3 and got:

(x+1)3(x+1)^2 + (x+1)^3 ?

Reply 895

Kevin De Bruyne

Original post by kiiten

Ah but im not sure about using the product rule with brackets. I tried it with (x+1)^3 and got:

(x+1)3(x+1)^2 + (x+1)^3 ?

The product rule is when you have two functions multiplying eachother - you're mixing it up with the chain rule a bit, though you do have to (spoiler alert) use the chain rule here.

If you say f(x) = (x+1)^3 and g(x) = (x+2)^6.. you do the rest :smile:

Reply 896

kiiten

Original post by SeanFM

I think thats what i did?

I made u = x+1 then v = u^3
then using the product rule i got the equation from my other post. Unless you make u = f(x) and v = g(x)?

Reply 897

Kevin De Bruyne

Original post by kiiten

I think thats what i did?

I made u = x+1 then v = u^3
then using the product rule i got the equation from my other post. Unless you make u = f(x) and v = g(x)?

I would suggest revisitng the product rule :smile:

What you have mentioned in your post there is the chain rule, which is used in this question but you have to apply to product rule first.

If you are ever stuck with something, check the definitions of whatever you're using then look for simple examples and understand them, then look for similar examples. :smile:

(edited 7 years ago)

Reply 898

kiiten

Original post by SeanFM

I would suggest revisitng the product rule :smile:

What you have mentioned in your post there is the chain rule, which is used in this question but you have to apply to product rule first.

Yeah I agree :biggrin:

Is the product rule uv^1 + vu^1 ?
If so then i made u = x+1 and v = u^3

Reply 899

Kevin De Bruyne

Original post by kiiten

Yeah I agree :biggrin:

Is the product rule uv^1 + vu^1 ?
If so then i made u = x+1 and v = u^3

I think you know what you mean but you're not typing / expressing it correctly. If u and v are functions, (uv)' = u'v + v'u is the product rule.

But with that second line you are defining the chain rule.. those are two different things. Again, I would suggest looking up the two to refresh your memory :borat: