2022/039) Find x integer such that $\frac{8^x - 2^x }{6^x - 3^x} = 2$

 We have numerator = $8^x - 2^x = (2^3)^x - 2^x = (2^(3x) - 2^x = 2^x(2^{2x} - 1)= 2^x(2^x + 1)(2^x-1)$

Denominator = $6^x - 2^x = 3^x(2^x-1)$

So $\frac{8^x - 2^x }{6^x - 3^x} = \frac{2^x(2^x+1)}{3^x} = 2$

Or $2^x(2^x+1) = 2 * 3^x$

x cannot be zero as in the original question x = 0 is not possible

So we must ahve $2^x = 2$ and $2^x+ 1 = 3^x$ by equating even on both sides and odd on both sides

1st equuation gives x = 1 and it satisfies 2nd equation as well.

so Solution x = 1