# Homework Help: Problem involving exponents

1. Jan 28, 2012

### Rron

A very logical problem.

1. The problem statement, all variables and given/known data
x5y=3z and x3y=z2-3z-2 so
what is (z)xy

2. Relevant equations
No equations, just multiplication,division etc.

3. The attempt at a solution
let xy=a
a^5=3z we can write this as a^15=(3z)^3
a^3=z^2-3z-2 also this a^15=(z^2-3z-2) and then equalize but I don't think I will get something, isn't it?

Last edited: Jan 28, 2012
2. Jan 28, 2012

### engphy2

try using 'ln' on both equations then substitution then binomial expansion.

3. Feb 2, 2012

### ehild

Re: A very logical problem.

Hi Rron,

It is a challenging problem. I did not find any nice solution. If you substitute z=(a^5)/3 into the second equation, you get an equation for a. Use wolframalpha.com to get the roots, choose the positive one.

ehild

4. Feb 3, 2012

### coolul007

since x^y = (3z)^(1/5), aren't we done, (z)^x^y = (z)^((3z)^(1/5))

5. Feb 3, 2012

### ehild

No, you have to determine z.

ehild

6. Feb 4, 2012

### coolul007

let a =x^y, then z =(a^5)/3, substituting in next equation; a^10 - 9a^5 - 9a^3 - 18 = 0, if we solve for a then z follows, etc. I do not have a program to do that so I'm stuck.

7. Feb 5, 2012

### ehild

The equation does not have a nice solution. You need a numerical one. wolframalpha.com helps.

ehild

8. Feb 5, 2012

### coolul007

When i numerically crunch it there are 2 real solutions 4 pi and -pi or something very close

9. Feb 7, 2012

### ehild

Are you sure? Have you tried to substitute them back?

ehild

10. Feb 7, 2012

### coolul007

Here is what Mathematica had to say:
{a -> -1.37397 - 0.905875i, z -> 3.9207- 0.906134i},
{a -> -1.37397 + 0.905875i, z -> 3.9207+ 0.906134i},
{a -> 1.68316, z -> 4.50308},
{a -> 0.543688+ 1.43263i, z -> 2.73342- 0.676945i},
{a -> 0.543688- 1.43263i, z -> 2.73342+ 0.676945i},
{a -> 0.785766+ 0.723955i, z -> -0.387992 - 0.254645i},
{a -> 0.785766- 0.723955i, z -> -0.387992 + 0.254645i},
{a -> -0.303313 + 1.18312i, z -> -0.86116 + 0.28155i},
{a -> -0.303313 - 1.18312i, z -> -0.86116 - 0.28155i},
{a -> -0.987501, z -> -0.313017}

11. Feb 8, 2012

### ehild

So what is z(xy)?

Was there any restriction for x,y,z? Are they real or complex numbers?

ehild

12. Feb 8, 2012

### coolul007

I have no idea if there were restrictions so here is the positve real answer:

since a = x^y, merely take all of these z's and raise them to the a power e.g.,

(4.50308)^(1.68316) = 12.588133023091346014101552100309

13. Feb 9, 2012

### ehild

It is all right, but do not write so many digits.

ehild