Best way to solve $24^{100} \times 1.5^{50} \times 12^{-149}$

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I think I could solve this but I would like to know the best way to do it with the least amount of calculations

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$$ 24^{100} \times 1.5^{50} \times 12^{-149} = (3\cdot 2^3)^{100} \times (3\cdot 2^{-1})^{50} \times (3 \cdot 2^2)^{-149} \\ = 3^{100} \cdot 2^{300}\cdot 3^{50}\cdot 2^{-50}\cdot 2^{-298}\cdot 3^{-149} = ...$$

You can add exponents to get the final answer

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As suggested by Mufasa in comments, expand in powers of $2$ and $3$: $$ \overbrace{2^{300}3^{100}}^{24^{100}}\overbrace{3^{50}2^{-50}}^{1.5^{50}}\overbrace{2^{-298}3^{-149}}^{12^{-149}}=3\cdot2^{-48} $$ We can also reduce computation with $2^{-48}=\left(\left(\left(\left(\frac18\right)^2\right)^2\right)^2\right)^2$