> since powers of 2 and 3 are always somewhat far apart.
This is the part where your argument fails. On a logarithmic scale, powers of 2 and powers of 3 can become arbitrarily close together.
To see this, notice that for any ε > 0, we can create a rational approximation x/y such that 0 < ln 2/ln 3 - x/y ≤ ε. Multiplying by y ln 3, we have 0 < y ln 2 - x ln 3 ≤ εy ln 3, or equivalently, x ln 3 < y ln 2 ≤ (x + εy) ln 3. Finally, we take the exponential to get 3^x < 2^y ≤ 3^(x + εy) = 3^(x(1 + εy/x)). As a consequence of Dirichlet's approximation theorem, εy/x can become arbitrarily small, so our upper bound 3^(x(1 + εy/x)) can become arbitrarily close to 3^x on a logarithmic scale.
This is the part where your argument fails. On a logarithmic scale, powers of 2 and powers of 3 can become arbitrarily close together.
To see this, notice that for any ε > 0, we can create a rational approximation x/y such that 0 < ln 2/ln 3 - x/y ≤ ε. Multiplying by y ln 3, we have 0 < y ln 2 - x ln 3 ≤ εy ln 3, or equivalently, x ln 3 < y ln 2 ≤ (x + εy) ln 3. Finally, we take the exponential to get 3^x < 2^y ≤ 3^(x + εy) = 3^(x(1 + εy/x)). As a consequence of Dirichlet's approximation theorem, εy/x can become arbitrarily small, so our upper bound 3^(x(1 + εy/x)) can become arbitrarily close to 3^x on a logarithmic scale.