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New Maths GCSE 9 -1 Exam Style Question, NEED HELP!!!

Hey Everyone,
I was doing some revision and I stumbled on this question. I have never seen an unknown in an indices. I know how to rearrange for x and all of that (cuz im in yr11..duuh) lol anyways the question is:
3^2x = 1/81

Any help?

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Reply 1
Avatar for Notnek
Notnek
21
Original post by NothingButWaleed
Hey Everyone,
I was doing some revision and I stumbled on this question. I have never seen an unknown in an indices. I know how to rearrange for x and all of that (cuz im in yr11..duuh) lol anyways the question is:
3^2x = 1/81

Any help?

At GCSE, you can't rearrange to solve these.

Start with 3x=813^x = 81. Can you see what xx must be here?

Then what about 3x=1813^x = \frac{1}{81}. Using your answer above, what do you think xx must be?

Finally can you use that to answer your question?

Please post all your thoughts / working if you get stuck.
(edited 7 years ago)
Original post by notnek
At GCSE, you can't rearrange to solve these.

Start with 3x=813^x = 81. Can you see what xx must be here?

Then what about 3x=1813^x = \frac{1}{81}. Using your answer above, what do you think xx must be?

Finally can you use that to answer your question?

Please post all your thoughts / working if you get stuck.


OK i understand but then it says 2x?
Look up logarithms
Reply 4
Avatar for Notnek
Notnek
21
Original post by NothingButWaleed
OK i understand but then it says 2x?

E.g. if you had 22x=162^{2x}=16

Since the solution to 2x=162^x = 16 is 44, for the above equation you need 2x=42x=4 so x=2x=2.

Can you try that for your question?
Reply 5
Avatar for Notnek
Notnek
21
Original post by Danielelliston
Look up logarithms

They're not part of GCSE. This question is meant to be done by considering powers of 3.
Original post by NothingButWaleed
Hey Everyone,
I was doing some revision and I stumbled on this question. I have never seen an unknown in an indices. I know how to rearrange for x and all of that (cuz im in yr11..duuh) lol anyways the question is:
3^2x = 1/81

Any help?


re write the rhs with a base of 3 to start with
OHHHHHH!!!!,

So if there is a 3, i need to do 3^4 to get 81, but if there is a 2x i need an integer which will give me 4 when multiplied by 2, that number is 2
But to get a fraction I need a negative indices so I will need -2

Hence,
3^2(-2) = 1/81

There!
Reply 8
Avatar for Zxrxh
Zxrxh
3
Original post by notnek
At GCSE, you can't rearrange to solve these.

Start with 3x=813^x = 81. Can you see what xx must be here?

Then what about 3x=1813^x = \frac{1}{81}. Using your answer above, what do you think xx must be?

Finally can you use that to answer your question?

Please post all your thoughts / working if you get stuck.


how do u use the different maths symbols?
Original post by Zxrxh
how do u use the different maths symbols?


use "latex" tags to get the different maths symbols...check this website out to see how to use them:
https://www.thestudentroom.co.uk/wiki/LaTex

12\frac{1}{2}
(edited 7 years ago)
Original post by NothingButWaleed
use "latex" tags to get the different maths symbols...check this website out to see how to use them:
https://www.thestudentroom.co.uk/wiki/LaTex

12\frac{1}{2}


To add to this, you can click on the expressions to see what code was used to format them :smile:
Reply 11
Avatar for Zxrxh
Zxrxh
3
Original post by NothingButWaleed
use "latex" tags to get the different maths symbols...check this website out to see how to use them:
https://www.thestudentroom.co.uk/wiki/LaTex

12\frac{1}{2}


thnxx
Reply 12
Avatar for Zxrxh
Zxrxh
3
How do you make t the subject of the formula v = u +at.
Original post by Zxrxh
How do you make t the subject of the formula v = u +at.


ok..so basically we know that at = a*t therefore to isolate t we would need to divide both sides of the equation by a..giving us
va=u+t\frac{v}{a} = u+t

then we would need to subtract "u" from both sides to, again, isolate "t" so we would get the new equation:

t=va−ut = \frac{v}{a} - u
Reply 14
Avatar for Zxrxh
Zxrxh
3
Original post by NothingButWaleed
ok..so basically we know that at = a*t therefore to isolate t we would need to divide both sides of the equation by a..giving us
va=u+t\frac{v}{a} = u+t

then we would need to subtract "u" from both sides to, again, isolate "t" so we would get the new equation:

t=va−ut = \frac{v}{a} - u


oh yeah thanks, that was like a grade 3 :colondollar:
Original post by Zxrxh
oh yeah thanks, that was like a grade 3 :colondollar:


lol yh no worries
Original post by NothingButWaleed
ok..so basically we know that at = a*t therefore to isolate t we would need to divide both sides of the equation by a..giving us
va=u+t\frac{v}{a} = u+t

then we would need to subtract "u" from both sides to, again, isolate "t" so we would get the new equation:

t=va−ut = \frac{v}{a} - u


You didn't divide both sides by aa correctly.

Original post by Zxrxh
oh yeah thanks, that was like a grade 3 :colondollar:
Reply 17
Avatar for Zxrxh
Zxrxh
3
Original post by RDKGames
You didn't divide both sides by aa correctly.


so what would the answer be ??
Reply 18
Avatar for Zxrxh
Zxrxh
3
im failing maths 100%
Original post by Zxrxh
so what would the answer be ??


v=u+atv=u+at

Dividing both sides a:

va=u+ata\frac{v}{a}=\frac{u+at}{a}

Split the fraction on the RHS into 2 different ones with the same denominator. Simplify where appropriate. Rearrange.

Have a go.

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