Solve the equation 4^(x/y+y/x)=32

Question

This question did not mention the need of the answer. Can it be found in finding x and y.

Answer 1

We can't solve this equation for a unique solution, however, we can identify the solution set as follows.

 

Let u = x/y. Then y/x = 1/u and substituting gives

 

4^(x/y+y/x) = 32

 

2^2x/y+2y/x = 2⁵

 

2x/y + y/x = 5

 

2u + 2/u = 5

 

2u² - 5u + 2 = 0

 

(2u - 1)(u - 2) = 0

 

u = 1/2 or u = 2

 

y/x = 2 or x/y = 2

 

So the solution set is two lines through the origin, one with slope 2 and the other with slope 1/2, but with the point (0,0) removed (you can't divide by 0).

Answer 2

4^(x/y+y/x)=32 can be written as 2^[2(x/y+y/x)] =2⁵

and by equating exponents, since the base is 2 for each of these exponentials, you have

2(x/y+y/x) = 5

and of course the left side is begging for use of the common denominator.

 

Ultimately, you'll have an equation equating an expression involving BOTH x & y equal to a number.

 

But two equations are normally needed to solve for two variables. So it appears to me that there might be more instructions (givens) than you have mentioned; until then, you're stymied.