# Logarithms and exponential functionsWatch

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#1
Solve the equation 8(8^(x-1) -1) = 7(4^x -2^(x+1))
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#2
(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
Last edited by Shas72; 1 week ago
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1 week ago
#3
(Original post by Shas72)
I tried to do it using cubic equation but it is not correct
TBH, a couple of solutions are fairly easy to spot.
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#4
(Original post by mqb2766)
TBH, a couple of solutions are fairly easy to spot.
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1 week ago
#5
(Original post by Shas72)
Looks ok so far. Is there something missing at the bottom?
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#6
(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
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1 week ago
#7
(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).
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#8
(Original post by mqb2766)
Looks ok so far. Is there something missing at the bottom?
I don't know how to go ahead
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1 week ago
#9
(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.
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#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
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1 week ago
#11
(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.
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#12
(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
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#13
(Original post by Shas72)
I did not get you.pls explain
So if take 1, 2,4 and take log
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1 week ago
#14
(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.
Last edited by mqb2766; 1 week ago
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1 week ago
#15
(Original post by Shas72)
So if take 1, 2,4 and take log
Take logs if necessary, but the answers are fairly trivial.
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#16
(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.
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1 week ago
#17
(Original post by Shas72)
So I get 1, 2,4 and then when I take their logs I get 0, 1,2.
yes
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#18
Thanks a lottt. I was thinking of factorising cubic equation completely. Now I understood. Thanks a lot again
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1 week ago
#19
(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.
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#20
(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
Last edited by Shas72; 1 week ago
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