Can anyone help me with this:

3^(n-6) x y^(4n-8) = 1

Find n and y.

Thanks so much!

There are three variables and I don't know how to solve it.

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- Feb 10th 2006, 06:27 PMardnasSolve algebra
Can anyone help me with this:

3^(n-6) x y^(4n-8) = 1

Find n and y.

Thanks so much!

There are three variables and I don't know how to solve it. - Feb 10th 2006, 07:43 PMticbolQuote:

Originally Posted by**ardnas**

Here is one way.

3^(n-6) * y^(4n-8) = 1

Divide both sides by 3^(n-6),

y^(4n-8) = 1/ [3^(n-6)]

y^(4n-8) = 3^[-(n-6)]

y^(4n-8) = 3^(6-n) -----------***

So, if (4n-8) = (6-n), then y=3.

4n -8 = 6 -n

4n +n = 6 +8

5n = 14

n = 14/5 = 2.8

Therefore, n=2.8 and y=3. ---------answer.

-----------------------------

There could be infinitely many sets of answer, not only (n=2.8, y=3), but I won't investigate that anymore. I assumed you need at least one set of answers only.

-------------------------

Check,

3^(n-6) * y^(4n-8) = 1

3^(2.8 -6) * 3^(4*2.8 -8) =? 1

3^(-3.2) * 3^(11.2 -8) =? 1

3^(-3.2) * 3^(3.2) =? 1

3^[(-3.2) +3.2] =? 1

3^0 =? 1

1 =? 1

Yes, so, OK. - Feb 10th 2006, 07:46 PMardnas
Oh I am sorry, the equation should read

3^(n-6) * y(4n-8) = a - Feb 10th 2006, 07:57 PMticbolQuote:

Originally Posted by**ardnas**

- Feb 10th 2006, 08:03 PMardnas
I am really sorry I mixed up the number and the letter.

I've tried using log but it just gets worse, it becomes more and more complicated. - Feb 10th 2006, 11:38 PMCaptainBlackQuote:

Originally Posted by**ardnas**

help. You won't get an answer like y=7, n=2, a=3 from this, as a single

equation will normally only give explicit numerical answers for one unknown.

It would be usual to treat "a" as an unknown constant and solve for the

other unknowns in terms of it, but here we still have two unknowns y and n,

which define a curve in the y n plane.

RonL