+0  
 
0
99
1
avatar

Given that log36 m = log42 n = log49(3m + 4n), compute the sum of all possible values of n/m

if m and n are both positive real numbers.

 Mar 11, 2023
 #1
avatar
0

Using the change of base formula for logarithms, we can rewrite the given equations as:

log m / log 36 = log n / log 42 = log (3m + 4n) / log 49

Let's start by equating the first two expressions:

log m / log 36 = log n / log 42

log m log 42 = log n log 36

log (m^log 42) = log (n^log 36)

m^log 42 = n^log 36

Taking the log of both sides with base 42, we get:

log 42 (m^log 42) = log 42 (n^log 36)

log 42 m = log 36 n

Using this result, we can write:

log m / log 36 = log n / log 42 = log (3m + 4n) / log 49

log (3m + 4n) / log 49 = log m / log 36 = log n / log 42

log (3m + 4n) = log 49 (m^(log 36) ) = log 49 (n^(log 42) )

3m + 4n = 49^(log 36) * m = 49^(log 42) * n

Using the relationship between logarithms and exponents, we can write:

49^(log 36) = 6^2^(2 log 3) = 6^4 log 3

49^(log 42) = 7^2^(2 log 3/2) = 7^2^(log 9) = 7^2

Substituting these values, we get:

3m + 4n = 6^4 log 3 * m = 49n

Simplifying, we get:

(6^4 log 3 - 4) n = 3m

Therefore, the possible values of n/m are given by:

n/m = (3m) / [(6^4 log 3 - 4) m] = 3 / (6^4 log 3 - 4)

Since there is only one possible value of n/m, we can directly compute it:

n/m = 3 / (6^4 log 3 - 4) = 3/12 = 1/4.

 Mar 11, 2023

3 Online Users

avatar