MichaelTait
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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?
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brianeverit
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(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
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GingerCodeMan
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Spoiler:
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 32^{v+4} = 8^{v^2-1}

 \left(2^5\right)^{v+4} = \left(2^3\right)^{v^2-1}

 2^{5v+20} = 2^{3v^2 - 3}

 5v + 20 = 3v^2 - 3

 3v^2 - 5v - 23 = 0

 v = \frac{5+\sqrt{301}}{6}, \frac{5 - \sqrt{301}}{6}

Hope that helps

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TenOfThem
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(Original post by GingerCodeMan)
Spoiler:
Show

 32^{v+4} = 8^{v^2-1}

 \left(2^5\right)^{v+4} = \left(2^3\right)^{v^2-1}

 2^{5v+20} = 2^{3v^2 - 3}

 5v + 20 = 3v^2 - 3

 3v^2 - 5v - 23 = 0

 v = \frac{5+\sqrt{301}}{6}, \frac{5 - \sqrt{301}}{6}

Hope that helps


Are you aware that full solutions are discouraged
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GingerCodeMan
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I was not aware. Why is that?
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GingerCodeMan
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(Original post by TenOfThem)
Are you aware that full solutions are discouraged
I was not aware. Why is that?
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TenOfThem
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(Original post by GingerCodeMan)
I was not aware. Why is that?
http://www.thestudentroom.co.uk/showthread.php?t=403989
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GingerCodeMan
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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.
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TenOfThem
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(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
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