1. ## Solve 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.

2. Originally Posted by ardnas
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.
I am sure that the "x" is not a variable here, that x is "times".

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

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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.

3. Oh I am sorry, the equation should read

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

4. Originally Posted by ardnas
Oh I am sorry, the equation should read

3^(n-6) * y(4n-8) = a
Zeez. Are you sure it is not equal to b instead?

5. 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.

6. Originally Posted by ardnas
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.
If you could tell us what you expect a solution to look like that might
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