Question

log base a 2 = m and logbase a 5 = n where 0 < a is not equal to 1 then log base a 500 =

Answer 1

Answer:

2m + 3n

Step-by-step explanation:

$log_{a}(2) \: = \: m \\ log_{a}(5) \: = \: n \\ find \: log_{a}(500) \: = \: ? \\ \\ log_{a}(500) \: = \: log_{a}(5 \times 5 \times 5 \times 2 \times 2) \\ log_{a}(500) \: = \: log_{a}( {5}^{3} \: \times \: {2}^{2} ) \\ log_{a}(500) \: = \: log_{a}( {5}^{3} ) \: + \: log_{a}( {2}^{2} ) \\ log_{a}(500) \: = \: 3 \: log_{a}(5) \: + \: 2 \: log_{a}(2) \\ log_{a}(500) \: = \: 3 \times (n) \: + \: 2 \times (m) \\ log_{a}(500) \: = \:3n \: + \: 2m$

Thanks!