1. **Problem statement:** Simplify the expression $\frac{5 \cdot 4^{15} \cdot 9^9 - 4 \cdot 3^{20} \cdot 8^9}{5 \cdot 2^9 \cdot 9^{16} - 7 \cdot 2^{29} \cdot 27^6}$. 2. **Recall important rules:** - Powers with the same base can be combined by addition or subtraction of exponents. - Factorization helps to simplify expressions. - $a^m \cdot a^n = a^{m+n}$. - $ (a^m)^n = a^{m \cdot n} $. 3. **Rewrite bases as powers of primes:** - $4 = 2^2$, $8 = 2^3$, $9 = 3^2$, $27 = 3^3$. 4. **Rewrite each term:** - Numerator first term: $5 \cdot (2^2)^{15} \cdot (3^2)^9 = 5 \cdot 2^{30} \cdot 3^{18}$. - Numerator second term: $4 \cdot 3^{20} \cdot (2^3)^9 = 4 \cdot 3^{20} \cdot 2^{27}$. - Denominator first term: $5 \cdot 2^9 \cdot (3^2)^{16} = 5 \cdot 2^9 \cdot 3^{32}$. - Denominator second term: $7 \cdot 2^{29} \cdot (3^3)^6 = 7 \cdot 2^{29} \cdot 3^{18}$. 5. **Rewrite numerator and denominator:** \[ \text{Numerator} = 5 \cdot 2^{30} \cdot 3^{18} - 4 \cdot 3^{20} \cdot 2^{27} \] \[ \text{Denominator} = 5 \cdot 2^9 \cdot 3^{32} - 7 \cdot 2^{29} \cdot 3^{18} \] 6. **Factor common terms in numerator:** - Common factor: $2^{27} \cdot 3^{18}$ \[ 2^{27} \cdot 3^{18} (5 \cdot 2^{3} - 4 \cdot 3^{2}) = 2^{27} \cdot 3^{18} (5 \cdot 8 - 4 \cdot 9) = 2^{27} \cdot 3^{18} (40 - 36) = 2^{27} \cdot 3^{18} \cdot 4 \] 7. **Factor common terms in denominator:** - Common factor: $2^{9} \cdot 3^{18}$ \[ 2^{9} \cdot 3^{18} (5 \cdot 3^{14} - 7 \cdot 2^{20}) \] 8. **Simplify denominator inside parentheses:** - $3^{14} = 4782969$ (large but exact), $2^{20} = 1048576$ - Calculate $5 \cdot 4782969 = 23914845$ - Calculate $7 \cdot 1048576 = 7340032$ - So inside parentheses: $23914845 - 7340032 = 16574813$ 9. **Rewrite denominator:** \[ 2^{9} \cdot 3^{18} \cdot 16574813 \] 10. **Rewrite numerator:** \[ 2^{27} \cdot 3^{18} \cdot 4 = 4 \cdot 2^{27} \cdot 3^{18} = 2^{2} \cdot 2^{27} \cdot 3^{18} = 2^{29} \cdot 3^{18} \cdot 4 \] Actually, since 4 is $2^2$, the numerator is $2^{29} \cdot 3^{18} \cdot 4$ is $2^{29} \cdot 3^{18} \cdot 4$ but we already counted 4 as $2^2$, so the numerator is $2^{29} \cdot 3^{18} \cdot 1$ (since 4 was factored out). 11. **Cancel common factors:** - Numerator: $2^{29} \cdot 3^{18}$ - Denominator: $2^{9} \cdot 3^{18} \cdot 16574813$ Divide numerator and denominator by $2^{9} \cdot 3^{18}$: $$ \frac{\cancel{2^{29}}^{20} \cdot \cancel{3^{18}}}{\cancel{2^{9}} \cdot \cancel{3^{18}} \cdot 16574813} = \frac{2^{20}}{16574813} $$ 12. **Final simplified form:** $$ \boxed{\frac{2^{20}}{16574813}} $$ 13. **Optional numeric approximation:** - $2^{20} = 1048576$ - So approximately $\frac{1048576}{16574813} \approx 0.0633$ **Answer:** $\frac{2^{20}}{16574813}$