1.3 Simplify: \[ \frac{2^{x}-2^{-x}}{4^{x}-1} \]

To simplify the expression \[ \frac{2^{x}-2^{-x}}{4^{x}-1}, \] we start by rewriting the denominator. Notice that \[ 4^x = (2^2)^x = (2^x)^2. \] So we can rewrite the denominator \(4^{x} - 1\) as \[ (2^x)^2 - 1. \] This expression is a difference of squares, which can be factored: \[ (2^x)^2 - 1 = (2^x - 1)(2^x + 1). \] Next, we simplify the numerator \(2^{x} - 2^{-x}\): \[ 2^{-x} = \frac{1}{2^x}, \] so we can express the numerator as \[ 2^{x} - \frac{1}{2^x} = \frac{2^{2x} - 1}{2^x}. \] Now, we can substitute this into our original expression: \[ \frac{\frac{2^{2x}-1}{2^x}}{(2^x - 1)(2^x + 1)}. \] This gives us: \[ \frac{2^{2x}-1}{2^x \cdot (2^x - 1)(2^x + 1)}. \] Again, we recognize \(2^{2x} - 1\) as another difference of squares, so we can factor: \[ 2^{2x} - 1 = (2^x - 1)(2^x + 1). \] Now we substitute this back into the expression: \[ \frac{(2^x - 1)(2^x + 1)}{2^x \cdot (2^x - 1)(2^x + 1)}. \] When we simplify this, we see that the \((2^x - 1)(2^x + 1)\) terms in the numerator and denominator cancel each other: \[ \frac{1}{2^x}. \] Thus, the simplified form is: \[ \boxed{\frac{1}{2^x}}. \]