Each of M and N is a positive real number that satisfies the following simultaneous equations:

• log₁₄(7) = M
• 14^N = 5

Determine the value of log₃₅(28) in terms of M and N.

Answer:   log(28,base=35) = (2-M) / (M+N)

M = log(7,base=14)
N = log(5,base=14)
X = log(28,base=35)
let a = ln2, b = ln5, c = ln7

M = c/(a + c)
N = b/(a + c)
X = (2a + c)/(b + c)

(M+N) = (b+c)/(a+c)
1/(M+N) = (a+c)/(b+c)
X+1 = (2a + 2c + b)/(b + c)
X+1 = 2(a+c)/(b+c) + (b)/(b + c)
X+1 = 2/(M+N) + (b)/(b + c)

but what is (b)/(b + c) ?
recall N = b/(a + c)
N /(M+N) = (b/(a + c)) * ((a+c)/(b+c)) = (b)/(b + c)

X+1 = 2/(M+N) + N /(M+N)
X = (2 + N) / (M+N)  - 1
X = (2 + N - M - N) / (M+N)
X = (2-M) / (M+N)

Posted by Larry on 2023-01-22 09:44:38