For equations involving $a^x$ and a^(-x), substitute t = $a^x$ (so a^(-x) = 1/t, t > 0). Example: 4^x - 62^x + 8 = 0. Write as (2^x)^2 - 6(2^x) + 8 = 0. Let t = 2^x: $t^2$ - 6t + 8 = 0, (t-2)(t-4) = 0, t = 2 or t = 4. So 2^x = 2 giving x = 1, or 2^x = 4 giving x = 2. Always discard negative values of t since $a^x$ > 0.