The Student Room Group

question on equations with powers

I have

a equation
6/5 = c^30,

c is a constant, how do i solve for C i forgot how to do these?

Scroll to see replies

So if you have anything to the power to solve for that anything you must do the inverse.

Inverse of squaring is the square root, inverse of cubing is the cube root.

Inverse of raising to the power of 30 you must "30th root" 1.2 (6/5).
Reply 2
Original post by plaguarist
So if you have anything to the power to solve for that anything you must do the inverse.

Inverse of squaring is the square root, inverse of cubing is the cube root.

Inverse of raising to the power of 30 you must "30th root" 1.2 (6/5).

Thanks, have another Q, how do i rearrange N = Ac^t to make T the subject?
Reply 3
Original post by SunDun111
Thanks, have another Q, how do i rearrange N = Ac^t to make T the subject?


Take logs.
Original post by SunDun111
Thanks, have another Q, how do i rearrange N = Ac^t to make T the subject?


What level/year you at/in?
Reply 5
Original post by britishtf2
What level/year you at/in?


A level, but am working on improving basic maths at the minute, it cost me a A last year
Original post by SunDun111
A level, but am working on improving basic maths at the minute, it cost me a A last year


Ah cool.

Use logs.
(edited 7 years ago)
Reply 7
Original post by britishtf2
Ah cool.

N = AC^T
log(N) = log(AC^T)
log(N) = T*log(AC)
T = log(N)/[log(AC)]


Full solutions are against forum guidelines.
Original post by Zacken
Full solutions are against forum guidelines.


Did not know that - thanks for the heads up.
Original post by SunDun111
A level, but am working on improving basic maths at the minute, it cost me a A last year


I couldn't recommend khan academy more, it is really good for the basics. It even goes well beyond that also, I used it this year for vector calculus, had I found out the did those kind of topics sooner I might have done a hell of a lot better!
Reply 10
Original post by britishtf2
Did not know that - thanks for the heads up.


There's a short posting guide at the top of the maths forum in red, might want to have a quick look at it. :smile:
Reply 11
Original post by britishtf2
Ah cool.

N = AC^T
log(N) = log(AC^T)
log(N) = T*log(AC)
T = log(N)/[log(AC)]


got that too straight away, was confused cause answer is 1/c multiplied by ln (n/a) = T
Original post by Zacken
There's a short posting guide at the top of the maths forum in red, might want to have a quick look at it. :smile:


Cheers :smile:
Reply 13
Original post by Zacken
There's a short posting guide at the top of the maths forum in red, might want to have a quick look at it. :smile:


I will post that M1 question today, that i got stuck on from a few days, think I explained the Q wrong.
Reply 14
Original post by SunDun111
got that too straight away, was confused cause answer is 1/c multiplied by ln (n/a) = T


Because that's the correct answer. The ^t is only on the C.

So log(AC^T) =/= T log (AC).
Original post by SunDun111
got that too straight away, was confused cause answer is 1/c multiplied by ln (n/a) = T


Actually, first line is as such:
N/A = C^T

Go from there.
Realised not all of the RHS is not ^T.
Reply 16
Original post by Zacken
Because that's the correct answer. The ^t is only on the C.

So log(AC^T) =/= T log (AC).


wouldnt you take log to the n thoguh as well, so it would be Log N / Log AC
Reply 17
Original post by britishtf2
Actually, first line is as such:
N/A = C^T

Go from there.
Realised not all of the RHS is not ^T.


Original post by Zacken
Because that's the correct answer. The ^t is only on the C.

So log(AC^T) =/= T log (AC).


Dm i figured it out, simply overcomplicated the question
Original post by Zacken
Because that's the correct answer. The ^t is only on the C.

So log(AC^T) =/= T log (AC).


Realised that as I read it again - I shouldn't within 20 mins of getting up xd
Reply 19
Original post by britishtf2
Realised that as I read it again - I shouldn't within 20 mins of getting up xd


I feel ya, just woken up as well. :lol:

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