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# Help! maths c 1! 20mins before its due watch

1. ive nevr encounted this kind of thing before help?? lool i have like 20mins before its due
( this sign '^' MEANS TO THE POWER)
a) given that y=2^x find expressions in terms of y for,
1)2^x+2

2)2^3-x

b) show that using y =2^x, the equation
2^x+2 +2^3-x=33 can be rewritten as 4y^2-33y+87=0

c) hence solve
2^x+2 +2^3-x=33

and lastly,
f(x)=x^1/2 -8x^-1/2
evaluate f(3) giving answer in simplest form with a rational denominator!

Helpp! im gonna be eaten alive!!!
2. We're here to help you with any problems you have understanding topics, not do your work for you because you left it to the last minute.
3. (Original post by cheesecake11)
ive nevr encounted this kind of thing before help?? lool i have like 20mins before its due
( this sign '^' MEANS TO THE POWER)
a) given that y=2^x find expressions in terms of y for,
1)2^x+2

2)2^3-x

b) show that using y =2^x, the equation
2^x+2 +2^3-x=33 can be rewritten as 4y^2-33y+87=0

c) hence solve
2^x+2 +2^3-x=33

and lastly,
f(x)=x^1/2 -8x^-1/2
evaluate f(3) giving answer in simplest form with a rational denominator!

Helpp! im gonna be eaten alive!!!
ai and aii equal to? ie 2^x+2 = 0
4. (Original post by Phenomenological)
We're here to help you with any problems you have understanding topics, not do your work for you because you left it to the last minute.
jeez sorry, i was vomitting all weekend and i got it on friday so i couldnt really do it in between vomitting and coughing my guts out!
5. (Original post by FranticMind)
ai and aii equal to? ie 2^x+2 = 0
i guess so, it doesnt say
6. (Original post by cheesecake11)
jeez sorry, i was vomitting all weekend and i got it on friday so i couldnt really do it in between vomitting and coughing my guts out!
Then tell the teacher that. Most of them do have the capacity for sympathy, though admittedly I've known a few who don't.
7. a) y=2^x so x=log (base2)y
And substitute.
i) is simple, but ii) is a logarithms question, I think, though b) implies it simplifies out.

b) Same

c) Solve equation for y (quadratic) and substitute=2^x

f(3) - just use a calculator with 3 in place of x.
8. (Original post by cheesecake11)
i guess so, it doesnt say
It doesn't need an equal to bit - it's just a term that they want in terms of y (see above).

Hope that helps in terms of method - I'll try working out the answers.
9. (Original post by cheesecake11)
i guess so, it doesnt say
Well its really simple substitution for a.i ? And then for a.ii you just rearrange the so you can substitute using logs.

For b again just substitute in and a hint would be use power laws on the 2^3-x term (a^3)/(a^x) = a^3-x. Then for c just simple quadratic that you made.

The last part is easy just put 3 in as the x and simplify.
10. (Original post by Octohedral)
a) y=2^x so x=log (base2)y
And substitute.
i) is simple, but ii) is a logarithms question, I think, though b) implies it simplifies out.

b) Same

c) Solve equation for y (quadratic) and substitute=2^x

f(3) - just use a calculator with 3 in place of x.
I wouldnt advise using a calculator, its best to do by hand as you can simplify in a more natural way. You could use it after for a messy value if you wanted though.
11. i got(i diddnt us ea cal) .-Sqrt6 +12 sqrt 3 divided by 12 for the function one?
12. (Original post by FranticMind)
I wouldnt advise using a calculator, its best to do by hand as you can simplify in a more natural way. You could use it after for a messy value if you wanted though.
Sorry, I agree with this.

OP;

a) i)y+2
ii) 2^3-x=2^3-log (2)y (2 is a subscript)

b) Are you sure you have copied a) and b) out correctly? I can't see how to eliminate the log, or why they would write '2^3' not 8.
13. (Original post by cheesecake11)
i got(i diddnt us ea cal) .-Sqrt6 +12 sqrt 3 divided by 12 for the function one?
I got sqrt(3)-8/sqrt(3), so -(5*sqrt(3))/3
14. (Original post by Octohedral)
I got sqrt(3)-8/sqrt(3), so -(5*sqrt(3))/3
oh? i guess i willl jst have to ask for an extention,i dont like rushing with maths!! arggh stupid illness! i 'll learn the logs thing and the function notation today and try again.thanks anyhow
15. (Original post by cheesecake11)
oh? i guess i willl jst have to ask for an extention,i dont like rushing with maths!! arggh stupid illness! i 'll learn the logs thing and the function notation today and try again.thanks anyhow
None of this requires logs as most log work uses a calculator and doesn't appear until C2

this should be a simple manipulation of laws of indices

a^m x a^n = a^(m + n)
a^m / a^n = a^(m - n)

if y=3^n

so 3^(3 + n) = 3^3 x 3^n = 27 x 3^n = 27y

and 3^(2 - n) = 3^2 / 3^n = 9/y

3 tips
Ask for help more than 20 mins before work should be in - most people who can help will be in lessons at 11:42 (either teaching or learning)
Post maths probs in the maths forum: http://www.thestudentroom.co.uk/forumdisplay.php?f=38

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