Question

If h^(3x-1) = k^(1-2y) and h^(2x-2) = k^(4y-3) , show that 16xy - 8y-11x + 5 = 0

Answer 1

Simultaneous linear equations in x and y enable x and y to be individually calculated in terms of h and k.

After so obtaining x and y, they can then by substituted into the equation containing just x and y to check whether it is true or false.

The complicated expressions involved in these calculations are prone to error, so some checking needs to be carried out before the proposed solution or proof can be validated.

(3x-1)log(h)=(1-2y)log(k); 3xlog(h)-log(h)=log(k)-2ylog(k);

(2x-2)log(h)=(4y-3)log(k); 2xlog(h)-2log(h)=4ylog(k)-3log(k);

3xlog(h)+2ylog(k)=log(hk); 6xlog(h)+4ylog(k)=log(h²k²);

2xlog(h)-4ylog(k)=log(h²/k³); 6xlog(h)-12ylog(k)=log(h⁶/k⁹);

8xlog(h)=log(h²k²)+log(h²/k³)=log(h⁴/k), x=log(h⁴/k)/(8log(h)).

16ylog(k)=log(h²k²)-log(h⁶/k⁹)=log(h^2-6k²⁺⁹)=log(k¹¹/h⁴),

y=log(k¹¹/h⁴)/(16log(k)).

16xy=16(log(h⁴/k)/(8log(h)))(log(k¹¹/h⁴)/(16log(k))),

16xy=log(h⁴/k)log(k¹¹/h⁴)/(8log(h)log(k));

8y=log(k¹¹/h⁴)/(2log(k));

11x=11log(h⁴/k)/(8log(h)).

16xy-8y-11x+5=

log(h⁴/k)log(k¹¹/h⁴)/(8log(h)log(k))-log(k¹¹/h⁴)/(2log(k))-11log(h⁴/k)/(8log(h))+5=

[log(h⁴/k)log(k¹¹/h⁴)-4log(h)log(k¹¹/h⁴)-11log(k)log(h⁴/k)+40log(h)log(k)]/(8log(h)log(k)).

This expression must be zero if 16xy-8y-11x+5=0 for all values of h and k. But it would appear that h and k are not independent and therefore the equation cannot be proved. However, it can theoretically be reduced to an equation relating h and k. An obvious solution is h=k=1, which satisfies the initial given equations for h and k, because log(h)=log(k)=0. Nevertheless these are constraints on h and k, and x and y cannot be defined as derived. The graph below shows how h (horizontal axis) and k (vertical axis) are related (red curve). The blue crosswires show the special case of h=k=1. The graph also shows that the curve is discontinuous.

These initial conditions can reduce to:

3x-1=1-2y; 2x-2=4y-3;

3x+2y=2, 6x+4y=4;

2x-4y=-1.

8x=3, x=⅜, 4y=2x+1, y=7/16.

16xy-8y-11x+5=21/8-7/2-33/8+5=(21-28-33+40)/8=0.

A SOLUTION: x=⅜, y=7/16, h=k=1.

Since h=k=1 then h^3x-1=k^1-2y and h^2x-2=k^4y-3 are true for all exponents, which means that, for example, x=y=0 would also satisfy these equations but not 16xy-8y-11x+5=0 which can only be satisfied when:

y=(11x-5)/[8(2x-1)] (for example, (⅜,7/16) is a solution).

TYPE OF GENERAL SOLUTION: y=(11x-5)/[8(2x-1)], h=k (when h=k the exponents are equal).

If k=h^λ then there are many more solutions (λ doesn't have to be an integer). λ=log(k)/log(h).

This produces h^3x-1=h^λ(1-2y) and h^2x-2=h^λ(4y-3), resulting in the simultaneous equations:

3x-1=λ(1-2y) and 2x-2=λ(4y-3), which can be solved for x and y:

3x-1=λ-2λy, 2x-2=4λy-3λ;

3x+2λy=λ+1,

2x-4λy=2-3λ,

6x+4λy=2λ+2,

8x=4-λ, x=(4-λ)/8;

(4-λ)/4-4λy=2-3λ,

4λy=(4-λ)/4-2+3λ=1-¼λ-2+3λ=(11λ-4)/4, y=(11λ-4)/(16λ).

In this form, the equations are a little more manageable.

16xy-8y-11x+5=(4-λ)(11λ-4)/(8λ)-(11λ-4)/(2λ)-11(4-λ)/8+5=

(44λ-16-11λ²+4λ-44λ+16-44λ+11λ²+40λ)/(8λ)=0, as required.

GENERAL SOLUTION

λ=log(k)/log(h); x=(4-λ)/8; y=(11λ-4)/(16λ). This solution is not a proof, because the given equation is only valid under certain conditions. It isn't generally true.