# Indicial Equations

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#1
Hello, I'm having a problem with a question set by a tutor, I just cant see where to go or where I am going wrong with it.

the questions states - solve the following equation:

32^(v+4) = 8^((v^2)-1)

(v+4)log32 = ((V^2)-1)log 8 - let log32=A and log8=B
(v+4)A = ((v^2)-1)B
Av + 4A = Bv^2 - B
so Bv^2-Av-B-4A = 0

but I cant get the quadratics formula to work, can anyone point me in the right direction?
0
6 years ago
#2
(Original post by MichaelTait)
Hello, I'm having a problem with a question set by a tutor, I just cant see where to go or where I am going wrong with it.

the questions states - solve the following equation:

32^(v+4) = 8^((v^2)-1)

(v+4)log32 = ((V^2)-1)log 8 - let log32=A and log8=B
(v+4)A = ((v^2)-1)B
Av + 4A = Bv^2 - B
so Bv^2-Av-B-4A = 0

but I cant get the quadratics formula to work, can anyone point me in the right direction?
Express both sides as a power of 2 and equate the powers to give you a simple quadratic in v
1
6 years ago
#3
1
6 years ago
#4

Are you aware that full solutions are discouraged
0
6 years ago
#5
I was not aware. Why is that?
0
6 years ago
#6
(Original post by TenOfThem)
Are you aware that full solutions are discouraged
I was not aware. Why is that?
0
6 years ago
#7
(Original post by GingerCodeMan)
I was not aware. Why is that?
0
6 years ago
#8
I just read a few threads about this, and I think in future I'll include hints and an explanation at the beginning of the post (in more spoiler tags), but if I were the OP, I would definitely appreciate getting a full worked solution of the question.

Thanks for the heads up though.
0
6 years ago
#9
(Original post by GingerCodeMan)
I just read a few threads about this, and I think in future I'll include hints and an explanation at the beginning of the post (in more spoiler tags)
That would be better

but if I were the OP, I would definitely appreciate getting a full worked solution of the question.
If I were the OP I would prefer to learn how to solve the problem so that I can solve others - almost inevitably the person asking for help with have a number of fully worked solutions (in the text book or from their lessons) that they can refer to - it is the ability to approach a question that they need help with

Thanks for the heads up though.
No problem
0
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