Logarithms question..

Question

Given log₂₅m=n, express log_m M²/125 in terms of n. 

Please help me

Answer 1

log₂₅M = (log_MM)/ log_M25 = 1 / log_M25 = n

 

So, log_M25 = 1/n

 

      log_M(5²) = 1/n

 

    2log_M5 = 1/n      log_M5 = 1/(2n)

 

Therefore, log_M[M²/125] = log_M(M²) - log_M(5³)

 

                                     = 2 - 3(log_M5)

 

                                     = 2 - 3(1/(2n))

 

                                     = (4n-3)/(2n)

 

Answer 2

log₂₅m=n

25ⁿ=m

log₅25ⁿ=log₅m

nlog₅25=log₅m

nlog₅5²=log₅m

2nlog₅5₌log₅m

2n=log₅m

 

log_m(m²/125)

use change of base formula

log₅(m²/125)/log₅m

log₅m²-log₅125

log₅m²-log₅5³

2log₅m-3log₅5

[2(2n)-3]/2n (from above-2n=log₅m)

(4n-3)/2n or 2-3/2n