The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

Logarithms and exponential functions

Solve the equation 8(8^(x-1) -1) = 7(4^x -2^(x+1))

Scroll to see replies

Reply 1
Avatar for Shas72
Shas72
OP
10
Original post by Shas72
Solve the equation 8(8^(x-1) -1) = 7(4^x -2^(x+1))

I tried to do it using cubic equation but it is not correct
(edited 4 years ago)
Reply 2
Avatar for mqb2766
mqb2766
21
Original post by Shas72
I tried to do it using cubic equation but it is not correct

Post your working?
TBH, a couple of solutions are fairly easy to spot.
Reply 3
Avatar for Shas72
Shas72
OP
10
Original post by mqb2766
Post your working?
TBH, a couple of solutions are fairly easy to spot.

1581604150608400183626978405584.jpg
Reply 4
Avatar for mqb2766
mqb2766
21
Original post by Shas72
1581604150608400183626978405584.jpg

Looks ok so far. Is there something missing at the bottom?
Reply 5
Avatar for Shas72
Shas72
OP
10
Original post by mqb2766
Looks ok so far. Is there something missing at the bottom?

No I tried solving the cubic equation by writing the factors of 8 but I didn't get the answer
Reply 6
Avatar for mqb2766
mqb2766
21
Original post by Shas72
No I tried solving the cubic equation by writing the factors of 8 but I didn't get the answer

Where is the u^3 cubic?
Why look for factors of 8 in the u cubic? Just solve for values of u, the take logs (assuming they're not trivial).
Reply 7
Avatar for Shas72
Shas72
OP
10
Original post by mqb2766
Looks ok so far. Is there something missing at the bottom?

I don't know how to go ahead
Reply 8
Avatar for mqb2766
mqb2766
21
Original post by Shas72
I don't know how to go ahead

Can you write the cubic in u? Its simple given the last line on your image.
Reply 9
Avatar for Shas72
Shas72
OP
10
Original post by mqb2766
Can you write the cubic in u? Its simple given the last line on your image.

So its u^3-7u^2+14u-8=0
Original post by Shas72
So its u^3-7u^2+14u-8=0

Yes. It is fairly easy to spot one or two values of u which make the left hand side zero (the roots). So you can write down one or two factors.
Just look at the coefficients and think what value(s) of u would make the terms sum to zero.
Reply 11
Avatar for Shas72
Shas72
OP
10
Original post by mqb2766
Where is the u^3 cubic?
Why look for factors of 8 in the u cubic? Just solve for values of u, the take logs (assuming they're not trivial).

I did not get you.pls explain
Reply 12
Avatar for Shas72
Shas72
OP
10
Original post by Shas72
I did not get you.pls explain

So if take 1, 2,4 and take log
Original post by Shas72
I did not get you.pls explain


When u = ?, when does the left hand side become zero. Try some values, then once you've found one, think about how it relates to the coefficients. That value is a root and (u-?) is a factor.
Once you have one factor, the cubic reduces to a quadratic which can then be easily solved.
(edited 4 years ago)
Original post by Shas72
So if take 1, 2,4 and take log


Take logs if necessary, but the answers are fairly trivial.
Reply 15
Avatar for Shas72
Shas72
OP
10
Original post by mqb2766
Yes. It is fairly easy to spot one or two values of u which make the left hand side zero (the roots). So you can write down one or two factors.
Just look at the coefficients and think what value(s) of u would make the terms sum to zero.

So I get 1, 2,4 and then when I take their logs I get 0, 1,2.
Original post by Shas72
So I get 1, 2,4 and then when I take their logs I get 0, 1,2.

yes
Reply 17
Avatar for Shas72
Shas72
OP
10
Thanks a lottt. I was thinking of factorising cubic equation completely. Now I understood. Thanks a lot again
Original post by Shas72
Thanks a lottt. I was thinking of factorising cubic equation completely. Now I understood. Thanks a lot again

You're unlikely to have to completely factorise a cubic. Often one root / factor will be fairly trivial and the problem will reduce to a quadratic.
Reply 19
Avatar for Shas72
Shas72
OP
10
Original post by mqb2766
You're unlikely to have to completely factorise a cubic. Often one root / factor will be fairly trivial and the problem will reduce to a quadratic.

Thanks
(edited 4 years ago)

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you