Question

3(log 5- log 3)-(log 5- 2log 6)=2-log n, find n​

Answer 1

Answer:

$\frac{100}{9 {e}^{2} }$

Step-by-step explanation:

$3( log(5) - log(3) ) - ( log(5) - 2 log(6) ) = 2 - log(n) \\ \\ 3( log( \frac{5}{3} ) ) - log( \frac{5}{12} ) = 2 - log(n) \\ \\ \\ log( \frac{125}{27} \times \frac{12}{5} ) = 2 - log(n) \\ \\ log( \frac{25 \times 4}{9} ) = 2 - log(n) \\ \\ log( \frac{25 \times 4}{9} ) - log(n) = 2 \\ \\ log( \frac{100}{9n} ) = 2 \\ \\ \frac{100}{9n} = {e}^{2 } \\ \\ n = \frac{100}{9 {e}^{2} }$