Question #cdbe1

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Question #cdbe1

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Answer 1 · 2016-12-24T14:34:30

$7 log (\frac{16}{15}) - 5 log (\frac{24}{25}) + 3 log (\frac{81}{80})$ $7log(16/15)-5log(24/25)+3log(81/80)$

$= {log (\frac{16}{15})}^{7} - {log (\frac{24}{25})}^{5} + {log (\frac{81}{80})}^{3}$ $=log(16/15)^7-log(24/25)^5+log(81/80)^3$

$= {log (\frac{2^{4}}{3 \times 5})}^{7} - {log (\frac{2^{3} \times 3}{5^{2}})}^{5} + {log (\frac{3^{4}}{2^{4} \times 5})}^{3}$ $=log(2^4/(3xx5))^7-log((2^3xx3)/5^2)^5+log(3^4/(2^4xx5))^3$

$= log (\frac{2^{28} \times 5^{10} \times 3^{12}}{3^{7} \times 5^{7} \times 2^{15} \times 3^{5} \times 2^{12} \times 5^{3}})$ $=log((2^28xx5^10xx3^12)/(3^7xx5^7xx2^15xx3^5xx2^12xx5^3))$

$= log 2$ $=log2$

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