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

Question

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

HINT: Express $24$ and $12$ as a product of their prime factors and express $1.5$ as $\frac{3}{2}$ then use standard laws of indices to simplify. — Mufasa, Oct 10 '15 at 10:19

score 0 · Accepted Answer · answered Oct 10 '15 at 10:32

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

You can add exponents to get the final answer

score 0 · Answer 2 · answered Oct 10 '15 at 10:35

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 $2^{-48}=\left(\left(\left(\left(\frac18\right)^2\right)^2\right)^2\right)^2$

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

2 Answers2