I don't see the relation

Let $a=2^x$ and $b = 2^y$ . We can then rewrite the given equations as $a - b = 1 \\ a^2 - b^2 = \frac{5}{3}$

Since $a - b = 1$ , we know that $(a - b)^2 = a^2 - 2ab + b^2 = 1$ . We can add this equation to $a^2 - b^2 = \frac{5}{3}$ to obtain

$2a^2 - 2ab = \frac{8}{3} \longrightarrow 2a(a - b) = \frac{8}{3} \longrightarrow a = \frac{4}{3} \longrightarrow b = \frac{1}{3}$

This means $2^x = \dfrac43$ and $2^y = \dfrac13$ . If we divide these two, we'll obtain $\dfrac{2^x}{2^y} = \frac{4/3}{1/3} \longrightarrow 2^{x - y} = 4$ , implying that $x - y = 2$ .

$2^{x}-2^{y}=1...(1)$

$2^{2x}-2^{2y}=\frac{5}{3}$

$\Rightarrow (2^{x}-2^{y})(2^{x}+2^{y})=\frac{5}{3}$

From (1),

$2^{x}+2^{y}=\frac{5}{3}...(2)$

From (1) & (2), Eliminating firstly $2^{y}$ & then $2^{x}$ , we get

$2*2^{x}=\frac{8}{3} \Rightarrow 2^x=\frac{4}{3}...(3)$

$2*2^{y}=\frac{2}{3} \Rightarrow 2^y=\frac{1}{3}...(4)$

$\frac{(3)}{(4)} \Rightarrow 2^{x-y}=4=2^2$

$\Rightarrow x-y =\boxed{2}$

:D

1st equation: 2^x - 2^y = 1

2nd equation: 4^x - 4^y = 5/3

Dissolve exponents in 2nd equation

4^x - 4^y = 5/3

2^(2x) - 2^(2y) = 5/3

Find exactly same terms in both equations. 1st equation has either 2^x or 2^y, and these variables can be seen in 2nd equation if exponents are dissolved enough.

2^(2x) - 2^(2y) = 5/3

(2^x)^2 - 2^(2y) = 5/3

Express 1st equation with respect to 2^x, (Note: can also be done with 2^y)

2^x - 2^y = 1

2^x = 1 + 2^y

Substitute 1 + 2^y into 2nd equation then simplify

(2^x)^2 - 2^(2y) = 5/3

(1+2^y)^2 - 2^(2y) = 5/3

1 + 2^y + 2^y + 2^(2y) - 2^(2y) = 5/3

1 + 2^y + 2^y = 5/3

1 + 2*2^y = 5/3

3/3 + 2*2^y = 5/3

2*2^y = 2/3

2^y = 2/3(1/2)

2^y = 1/3

Substitute 1/3 into 1st equation

2^x - 2^y = 1

2^x - 1/3 = 3/3

2^x = 4/3

Express 2^x and 2^y into log.

2^y = 1/3

log(2)(1/3) = y

2^x = 4/3

log(2)(4/3) = x

Substitute logs into x - y

x - y = log(2)(4/3) - log(2)(1/3) = log(2) [(4/3)/(1/3)] = log(2)(4) = 2

Some of the guys have done in the similar way I followed. There is nothing new.

4^{x} = 2^{2x}

• 4^{x}-4^{y} = (2^{x}-2^{y})*(2^{x}+2^{y})
• 2^{x}+2^{y} =1;
• x-y=log(2)(4/3) - log(2)(1/3) = 2

4^x=(2^2)^x =(2^x)^2,,,,,4^y=(2^y)^2,,,,,,,,,put 2^x=m&2^y=n,,,,,,,,,so 2^x-2^y=m-n ,,,4^x-4^y=m^2 -n^2 ....... . m-n=1& m^2 -n^2=(5÷3). * * m2 -(m -1)^2=5÷3. * . m ^2-m^2+2m-1=5÷3,,,,,,m =4÷3,,,,,,so n =1÷3,, * 2^x=4÷3,,,2^y=1÷3 by dividing 2 expression 2^x÷2^y=4,,,so 2^(x-y)=2^2,x-y =2######