0\\t^{2} + 2t - 8 = 0\\ t_{1} = 2; \ t_{2} \neq -4\\2^{x} = 2\\x = 1" alt="_{B}) \ 4^{x} + 2^{x+1} - 8 = 0\\2^{2x} + 2 \ \cdotp 2^{x} - 8 = 0\\t = 2^{x}, \ t > 0\\t^{2} + 2t - 8 = 0\\ t_{1} = 2; \ t_{2} \neq -4\\2^{x} = 2\\x = 1" align="absmiddle" class="latex-formula">