Question

7log base2 (16/15)+5logbase 2 (25/24)

Answer 1

Answer:

Given,

→ $7 log_{2}( \frac{16}{15} ) + 5 log_{2}( \frac{25}{24} )$

= $log_{2}(( { \frac{16}{15}) }^{7} ) + log_{2}(( { \frac{25}{24}) }^{5} )$

= $log_{2}( ( { \frac{16}{15} )}^{7} \times ( ({ \frac{25}{24}) }^{5} )$

= $log_{2}( \frac{{2}^{28} \times {5}^{10} }{ {2}^{15} \times {3}^{12} \times {5}^{7} } )$

= $log_{2}( \frac{ {2}^{13} \times {5}^{3} }{ {3}^{12} } )$

= $13 log_{2}(2) + 3 log_{2}(5) - 12 log_{2}(3)$

= $13 + 3 log_{2}(5) - 12 log_{2}(3)$

Formulae used:-

1) log(ab) = log(a) + log(b)

2) log(a/b) = log(a) - log(b)

3) log(a^(m)) = mlog(a)