Question

if 49^n+1*7^n-(343)n/7^3m * 2^4 = 3/343, prove that m=n+1

Answer 1

Step-by-step explanation:

Given If 49^n+1*7^n-(343)n/7^3m * 2^4 = 3/343, prove that m=n+1

• So we have the equation
•         49^n + 1 x 7^n – 343^n / 7^3m x 2^4 = 3 / 343
•        We can write this as  
•         (7^2)^n + 1 x 7^n – (7^3)^n / 7^3m x 2^4 = 3 / 7^3
•          We get by the law of exponents a^m x a^n = a^m + n
•           7^2n + 2 x 7^n – 7^3n / 7^3m x 2^4 = 3/7^3
•               7^2n + 2 + n – 7^3n / 7^3m x 48 = 1/7^3
•              7^3n + 2 – 7^3n / 7^3m x 48 = 1/7^3
•             7^3n x 7^2 – 7^3n / 7^3m x 48 = 1/7^3
•             7^3n (7^2 – 1) / 7^3m x 48 = 1/7^3
•             7^3n x 48 / 7^3m x 48 = 1/7^3
•            7^3n x 7^3 = 7^3m
•        Since bases are same exponents are equal.
•           So 3n + 3 = 3m
•             3(n + 1) = 3m
•           Or m = n + 1 (proved)

Reference link will be

https://brainly.in/question/21370309