Question

How would we solve this (7)^2x-4=1

Answer 1

Given
(7)^2x-4 = 1
Multiply both sides by 7 
[(7)^2x-4] × 7 = 1 × 7
⇒ 7^ (2x-4+1) = 7^1  [ a^n × a^m = a^(n+m)]
⇒ 7^ (2x-3) = 7^1
⇒ 2x-3 = 1                 [∵ Bases are same power are equal]
⇒ 2x = 1+3
⇒ 2x = 4
⇒ x = 4/2
∴ x = 2